1\documentclass[b5paper]{book} 2\usepackage{hyperref} 3\usepackage{makeidx} 4\usepackage{amssymb} 5\usepackage{color} 6\usepackage{alltt} 7\usepackage{graphicx} 8\usepackage{layout} 9\def\union{\cup} 10\def\intersect{\cap} 11\def\getsrandom{\stackrel{\rm R}{\gets}} 12\def\cross{\times} 13\def\cat{\hspace{0.5em} \| \hspace{0.5em}} 14\def\catn{$\|$} 15\def\divides{\hspace{0.3em} | \hspace{0.3em}} 16\def\nequiv{\not\equiv} 17\def\approx{\raisebox{0.2ex}{\mbox{\small $\sim$}}} 18\def\lcm{{\rm lcm}} 19\def\gcd{{\rm gcd}} 20\def\log{{\rm log}} 21\def\ord{{\rm ord}} 22\def\abs{{\mathit abs}} 23\def\rep{{\mathit rep}} 24\def\mod{{\mathit\ mod\ }} 25\renewcommand{\pmod}[1]{\ ({\rm mod\ }{#1})} 26\newcommand{\floor}[1]{\left\lfloor{#1}\right\rfloor} 27\newcommand{\ceil}[1]{\left\lceil{#1}\right\rceil} 28\def\Or{{\rm\ or\ }} 29\def\And{{\rm\ and\ }} 30\def\iff{\hspace{1em}\Longleftrightarrow\hspace{1em}} 31\def\implies{\Rightarrow} 32\def\undefined{{\rm ``undefined"}} 33\def\Proof{\vspace{1ex}\noindent {\bf Proof:}\hspace{1em}} 34\let\oldphi\phi 35\def\phi{\varphi} 36\def\Pr{{\rm Pr}} 37\newcommand{\str}[1]{{\mathbf{#1}}} 38\def\F{{\mathbb F}} 39\def\N{{\mathbb N}} 40\def\Z{{\mathbb Z}} 41\def\R{{\mathbb R}} 42\def\C{{\mathbb C}} 43\def\Q{{\mathbb Q}} 44\definecolor{DGray}{gray}{0.5} 45\newcommand{\emailaddr}[1]{\mbox{$<${#1}$>$}} 46\def\twiddle{\raisebox{0.3ex}{\mbox{\tiny $\sim$}}} 47\def\gap{\vspace{0.5ex}} 48\makeindex 49\begin{document} 50\frontmatter 51\pagestyle{empty} 52\title{Multi--Precision Math} 53\author{\mbox{ 54%\begin{small} 55\begin{tabular}{c} 56Tom St Denis \\ 57Algonquin College \\ 58\\ 59Mads Rasmussen \\ 60Open Communications Security \\ 61\\ 62Greg Rose \\ 63QUALCOMM Australia \\ 64\end{tabular} 65%\end{small} 66} 67} 68\maketitle 69This text has been placed in the public domain. This text corresponds to the v0.39 release of the 70LibTomMath project. 71 72\begin{alltt} 73Tom St Denis 74111 Banning Rd 75Ottawa, Ontario 76K2L 1C3 77Canada 78 79Phone: 1-613-836-3160 80Email: tomstdenis@gmail.com 81\end{alltt} 82 83This text is formatted to the international B5 paper size of 176mm wide by 250mm tall using the \LaTeX{} 84{\em book} macro package and the Perl {\em booker} package. 85 86\tableofcontents 87\listoffigures 88\chapter*{Prefaces} 89When I tell people about my LibTom projects and that I release them as public domain they are often puzzled. 90They ask why I did it and especially why I continue to work on them for free. The best I can explain it is ``Because I can.'' 91Which seems odd and perhaps too terse for adult conversation. I often qualify it with ``I am able, I am willing.'' which 92perhaps explains it better. I am the first to admit there is not anything that special with what I have done. Perhaps 93others can see that too and then we would have a society to be proud of. My LibTom projects are what I am doing to give 94back to society in the form of tools and knowledge that can help others in their endeavours. 95 96I started writing this book because it was the most logical task to further my goal of open academia. The LibTomMath source 97code itself was written to be easy to follow and learn from. There are times, however, where pure C source code does not 98explain the algorithms properly. Hence this book. The book literally starts with the foundation of the library and works 99itself outwards to the more complicated algorithms. The use of both pseudo--code and verbatim source code provides a duality 100of ``theory'' and ``practice'' that the computer science students of the world shall appreciate. I never deviate too far 101from relatively straightforward algebra and I hope that this book can be a valuable learning asset. 102 103This book and indeed much of the LibTom projects would not exist in their current form if it was not for a plethora 104of kind people donating their time, resources and kind words to help support my work. Writing a text of significant 105length (along with the source code) is a tiresome and lengthy process. Currently the LibTom project is four years old, 106comprises of literally thousands of users and over 100,000 lines of source code, TeX and other material. People like Mads and Greg 107were there at the beginning to encourage me to work well. It is amazing how timely validation from others can boost morale to 108continue the project. Definitely my parents were there for me by providing room and board during the many months of work in 2003. 109 110To my many friends whom I have met through the years I thank you for the good times and the words of encouragement. I hope I 111honour your kind gestures with this project. 112 113Open Source. Open Academia. Open Minds. 114 115\begin{flushright} Tom St Denis \end{flushright} 116 117\newpage 118I found the opportunity to work with Tom appealing for several reasons, not only could I broaden my own horizons, but also 119contribute to educate others facing the problem of having to handle big number mathematical calculations. 120 121This book is Tom's child and he has been caring and fostering the project ever since the beginning with a clear mind of 122how he wanted the project to turn out. I have helped by proofreading the text and we have had several discussions about 123the layout and language used. 124 125I hold a masters degree in cryptography from the University of Southern Denmark and have always been interested in the 126practical aspects of cryptography. 127 128Having worked in the security consultancy business for several years in S\~{a}o Paulo, Brazil, I have been in touch with a 129great deal of work in which multiple precision mathematics was needed. Understanding the possibilities for speeding up 130multiple precision calculations is often very important since we deal with outdated machine architecture where modular 131reductions, for example, become painfully slow. 132 133This text is for people who stop and wonder when first examining algorithms such as RSA for the first time and asks 134themselves, ``You tell me this is only secure for large numbers, fine; but how do you implement these numbers?'' 135 136\begin{flushright} 137Mads Rasmussen 138 139S\~{a}o Paulo - SP 140 141Brazil 142\end{flushright} 143 144\newpage 145It's all because I broke my leg. That just happened to be at about the same time that Tom asked for someone to review the section of the book about 146Karatsuba multiplication. I was laid up, alone and immobile, and thought ``Why not?'' I vaguely knew what Karatsuba multiplication was, but not 147really, so I thought I could help, learn, and stop myself from watching daytime cable TV, all at once. 148 149At the time of writing this, I've still not met Tom or Mads in meatspace. I've been following Tom's progress since his first splash on the 150sci.crypt Usenet news group. I watched him go from a clueless newbie, to the cryptographic equivalent of a reformed smoker, to a real 151contributor to the field, over a period of about two years. I've been impressed with his obvious intelligence, and astounded by his productivity. 152Of course, he's young enough to be my own child, so he doesn't have my problems with staying awake. 153 154When I reviewed that single section of the book, in its very earliest form, I was very pleasantly surprised. So I decided to collaborate more fully, 155and at least review all of it, and perhaps write some bits too. There's still a long way to go with it, and I have watched a number of close 156friends go through the mill of publication, so I think that the way to go is longer than Tom thinks it is. Nevertheless, it's a good effort, 157and I'm pleased to be involved with it. 158 159\begin{flushright} 160Greg Rose, Sydney, Australia, June 2003. 161\end{flushright} 162 163\mainmatter 164\pagestyle{headings} 165\chapter{Introduction} 166\section{Multiple Precision Arithmetic} 167 168\subsection{What is Multiple Precision Arithmetic?} 169When we think of long-hand arithmetic such as addition or multiplication we rarely consider the fact that we instinctively 170raise or lower the precision of the numbers we are dealing with. For example, in decimal we almost immediate can 171reason that $7$ times $6$ is $42$. However, $42$ has two digits of precision as opposed to one digit we started with. 172Further multiplications of say $3$ result in a larger precision result $126$. In these few examples we have multiple 173precisions for the numbers we are working with. Despite the various levels of precision a single subset\footnote{With the occasional optimization.} 174 of algorithms can be designed to accomodate them. 175 176By way of comparison a fixed or single precision operation would lose precision on various operations. For example, in 177the decimal system with fixed precision $6 \cdot 7 = 2$. 178 179Essentially at the heart of computer based multiple precision arithmetic are the same long-hand algorithms taught in 180schools to manually add, subtract, multiply and divide. 181 182\subsection{The Need for Multiple Precision Arithmetic} 183The most prevalent need for multiple precision arithmetic, often referred to as ``bignum'' math, is within the implementation 184of public-key cryptography algorithms. Algorithms such as RSA \cite{RSAREF} and Diffie-Hellman \cite{DHREF} require 185integers of significant magnitude to resist known cryptanalytic attacks. For example, at the time of this writing a 186typical RSA modulus would be at least greater than $10^{309}$. However, modern programming languages such as ISO C \cite{ISOC} and 187Java \cite{JAVA} only provide instrinsic support for integers which are relatively small and single precision. 188 189\begin{figure}[!here] 190\begin{center} 191\begin{tabular}{|r|c|} 192\hline \textbf{Data Type} & \textbf{Range} \\ 193\hline char & $-128 \ldots 127$ \\ 194\hline short & $-32768 \ldots 32767$ \\ 195\hline long & $-2147483648 \ldots 2147483647$ \\ 196\hline long long & $-9223372036854775808 \ldots 9223372036854775807$ \\ 197\hline 198\end{tabular} 199\end{center} 200\caption{Typical Data Types for the C Programming Language} 201\label{fig:ISOC} 202\end{figure} 203 204The largest data type guaranteed to be provided by the ISO C programming 205language\footnote{As per the ISO C standard. However, each compiler vendor is allowed to augment the precision as they 206see fit.} can only represent values up to $10^{19}$ as shown in figure \ref{fig:ISOC}. On its own the C language is 207insufficient to accomodate the magnitude required for the problem at hand. An RSA modulus of magnitude $10^{19}$ could be 208trivially factored\footnote{A Pollard-Rho factoring would take only $2^{16}$ time.} on the average desktop computer, 209rendering any protocol based on the algorithm insecure. Multiple precision algorithms solve this very problem by 210extending the range of representable integers while using single precision data types. 211 212Most advancements in fast multiple precision arithmetic stem from the need for faster and more efficient cryptographic 213primitives. Faster modular reduction and exponentiation algorithms such as Barrett's algorithm, which have appeared in 214various cryptographic journals, can render algorithms such as RSA and Diffie-Hellman more efficient. In fact, several 215major companies such as RSA Security, Certicom and Entrust have built entire product lines on the implementation and 216deployment of efficient algorithms. 217 218However, cryptography is not the only field of study that can benefit from fast multiple precision integer routines. 219Another auxiliary use of multiple precision integers is high precision floating point data types. 220The basic IEEE \cite{IEEE} standard floating point type is made up of an integer mantissa $q$, an exponent $e$ and a sign bit $s$. 221Numbers are given in the form $n = q \cdot b^e \cdot -1^s$ where $b = 2$ is the most common base for IEEE. Since IEEE 222floating point is meant to be implemented in hardware the precision of the mantissa is often fairly small 223(\textit{23, 48 and 64 bits}). The mantissa is merely an integer and a multiple precision integer could be used to create 224a mantissa of much larger precision than hardware alone can efficiently support. This approach could be useful where 225scientific applications must minimize the total output error over long calculations. 226 227Yet another use for large integers is within arithmetic on polynomials of large characteristic (i.e. $GF(p)[x]$ for large $p$). 228In fact the library discussed within this text has already been used to form a polynomial basis library\footnote{See \url{http://poly.libtomcrypt.org} for more details.}. 229 230\subsection{Benefits of Multiple Precision Arithmetic} 231\index{precision} 232The benefit of multiple precision representations over single or fixed precision representations is that 233no precision is lost while representing the result of an operation which requires excess precision. For example, 234the product of two $n$-bit integers requires at least $2n$ bits of precision to be represented faithfully. A multiple 235precision algorithm would augment the precision of the destination to accomodate the result while a single precision system 236would truncate excess bits to maintain a fixed level of precision. 237 238It is possible to implement algorithms which require large integers with fixed precision algorithms. For example, elliptic 239curve cryptography (\textit{ECC}) is often implemented on smartcards by fixing the precision of the integers to the maximum 240size the system will ever need. Such an approach can lead to vastly simpler algorithms which can accomodate the 241integers required even if the host platform cannot natively accomodate them\footnote{For example, the average smartcard 242processor has an 8 bit accumulator.}. However, as efficient as such an approach may be, the resulting source code is not 243normally very flexible. It cannot, at runtime, accomodate inputs of higher magnitude than the designer anticipated. 244 245Multiple precision algorithms have the most overhead of any style of arithmetic. For the the most part the 246overhead can be kept to a minimum with careful planning, but overall, it is not well suited for most memory starved 247platforms. However, multiple precision algorithms do offer the most flexibility in terms of the magnitude of the 248inputs. That is, the same algorithms based on multiple precision integers can accomodate any reasonable size input 249without the designer's explicit forethought. This leads to lower cost of ownership for the code as it only has to 250be written and tested once. 251 252\section{Purpose of This Text} 253The purpose of this text is to instruct the reader regarding how to implement efficient multiple precision algorithms. 254That is to not only explain a limited subset of the core theory behind the algorithms but also the various ``house keeping'' 255elements that are neglected by authors of other texts on the subject. Several well reknowned texts \cite{TAOCPV2,HAC} 256give considerably detailed explanations of the theoretical aspects of algorithms and often very little information 257regarding the practical implementation aspects. 258 259In most cases how an algorithm is explained and how it is actually implemented are two very different concepts. For 260example, the Handbook of Applied Cryptography (\textit{HAC}), algorithm 14.7 on page 594, gives a relatively simple 261algorithm for performing multiple precision integer addition. However, the description lacks any discussion concerning 262the fact that the two integer inputs may be of differing magnitudes. As a result the implementation is not as simple 263as the text would lead people to believe. Similarly the division routine (\textit{algorithm 14.20, pp. 598}) does not 264discuss how to handle sign or handle the dividend's decreasing magnitude in the main loop (\textit{step \#3}). 265 266Both texts also do not discuss several key optimal algorithms required such as ``Comba'' and Karatsuba multipliers 267and fast modular inversion, which we consider practical oversights. These optimal algorithms are vital to achieve 268any form of useful performance in non-trivial applications. 269 270To solve this problem the focus of this text is on the practical aspects of implementing a multiple precision integer 271package. As a case study the ``LibTomMath''\footnote{Available at \url{http://math.libtomcrypt.com}} package is used 272to demonstrate algorithms with real implementations\footnote{In the ISO C programming language.} that have been field 273tested and work very well. The LibTomMath library is freely available on the Internet for all uses and this text 274discusses a very large portion of the inner workings of the library. 275 276The algorithms that are presented will always include at least one ``pseudo-code'' description followed 277by the actual C source code that implements the algorithm. The pseudo-code can be used to implement the same 278algorithm in other programming languages as the reader sees fit. 279 280This text shall also serve as a walkthrough of the creation of multiple precision algorithms from scratch. Showing 281the reader how the algorithms fit together as well as where to start on various taskings. 282 283\section{Discussion and Notation} 284\subsection{Notation} 285A multiple precision integer of $n$-digits shall be denoted as $x = (x_{n-1}, \ldots, x_1, x_0)_{ \beta }$ and represent 286the integer $x \equiv \sum_{i=0}^{n-1} x_i\beta^i$. The elements of the array $x$ are said to be the radix $\beta$ digits 287of the integer. For example, $x = (1,2,3)_{10}$ would represent the integer 288$1\cdot 10^2 + 2\cdot10^1 + 3\cdot10^0 = 123$. 289 290\index{mp\_int} 291The term ``mp\_int'' shall refer to a composite structure which contains the digits of the integer it represents, as well 292as auxilary data required to manipulate the data. These additional members are discussed further in section 293\ref{sec:MPINT}. For the purposes of this text a ``multiple precision integer'' and an ``mp\_int'' are assumed to be 294synonymous. When an algorithm is specified to accept an mp\_int variable it is assumed the various auxliary data members 295are present as well. An expression of the type \textit{variablename.item} implies that it should evaluate to the 296member named ``item'' of the variable. For example, a string of characters may have a member ``length'' which would 297evaluate to the number of characters in the string. If the string $a$ equals ``hello'' then it follows that 298$a.length = 5$. 299 300For certain discussions more generic algorithms are presented to help the reader understand the final algorithm used 301to solve a given problem. When an algorithm is described as accepting an integer input it is assumed the input is 302a plain integer with no additional multiple-precision members. That is, algorithms that use integers as opposed to 303mp\_ints as inputs do not concern themselves with the housekeeping operations required such as memory management. These 304algorithms will be used to establish the relevant theory which will subsequently be used to describe a multiple 305precision algorithm to solve the same problem. 306 307\subsection{Precision Notation} 308The variable $\beta$ represents the radix of a single digit of a multiple precision integer and 309must be of the form $q^p$ for $q, p \in \Z^+$. A single precision variable must be able to represent integers in 310the range $0 \le x < q \beta$ while a double precision variable must be able to represent integers in the range 311$0 \le x < q \beta^2$. The extra radix-$q$ factor allows additions and subtractions to proceed without truncation of the 312carry. Since all modern computers are binary, it is assumed that $q$ is two. 313 314\index{mp\_digit} \index{mp\_word} 315Within the source code that will be presented for each algorithm, the data type \textbf{mp\_digit} will represent 316a single precision integer type, while, the data type \textbf{mp\_word} will represent a double precision integer type. In 317several algorithms (notably the Comba routines) temporary results will be stored in arrays of double precision mp\_words. 318For the purposes of this text $x_j$ will refer to the $j$'th digit of a single precision array and $\hat x_j$ will refer to 319the $j$'th digit of a double precision array. Whenever an expression is to be assigned to a double precision 320variable it is assumed that all single precision variables are promoted to double precision during the evaluation. 321Expressions that are assigned to a single precision variable are truncated to fit within the precision of a single 322precision data type. 323 324For example, if $\beta = 10^2$ a single precision data type may represent a value in the 325range $0 \le x < 10^3$, while a double precision data type may represent a value in the range $0 \le x < 10^5$. Let 326$a = 23$ and $b = 49$ represent two single precision variables. The single precision product shall be written 327as $c \leftarrow a \cdot b$ while the double precision product shall be written as $\hat c \leftarrow a \cdot b$. 328In this particular case, $\hat c = 1127$ and $c = 127$. The most significant digit of the product would not fit 329in a single precision data type and as a result $c \ne \hat c$. 330 331\subsection{Algorithm Inputs and Outputs} 332Within the algorithm descriptions all variables are assumed to be scalars of either single or double precision 333as indicated. The only exception to this rule is when variables have been indicated to be of type mp\_int. This 334distinction is important as scalars are often used as array indicies and various other counters. 335 336\subsection{Mathematical Expressions} 337The $\lfloor \mbox{ } \rfloor$ brackets imply an expression truncated to an integer not greater than the expression 338itself. For example, $\lfloor 5.7 \rfloor = 5$. Similarly the $\lceil \mbox{ } \rceil$ brackets imply an expression 339rounded to an integer not less than the expression itself. For example, $\lceil 5.1 \rceil = 6$. Typically when 340the $/$ division symbol is used the intention is to perform an integer division with truncation. For example, 341$5/2 = 2$ which will often be written as $\lfloor 5/2 \rfloor = 2$ for clarity. When an expression is written as a 342fraction a real value division is implied, for example ${5 \over 2} = 2.5$. 343 344The norm of a multiple precision integer, for example $\vert \vert x \vert \vert$, will be used to represent the number of digits in the representation 345of the integer. For example, $\vert \vert 123 \vert \vert = 3$ and $\vert \vert 79452 \vert \vert = 5$. 346 347\subsection{Work Effort} 348\index{big-Oh} 349To measure the efficiency of the specified algorithms, a modified big-Oh notation is used. In this system all 350single precision operations are considered to have the same cost\footnote{Except where explicitly noted.}. 351That is a single precision addition, multiplication and division are assumed to take the same time to 352complete. While this is generally not true in practice, it will simplify the discussions considerably. 353 354Some algorithms have slight advantages over others which is why some constants will not be removed in 355the notation. For example, a normal baseline multiplication (section \ref{sec:basemult}) requires $O(n^2)$ work while a 356baseline squaring (section \ref{sec:basesquare}) requires $O({{n^2 + n}\over 2})$ work. In standard big-Oh notation these 357would both be said to be equivalent to $O(n^2)$. However, 358in the context of the this text this is not the case as the magnitude of the inputs will typically be rather small. As a 359result small constant factors in the work effort will make an observable difference in algorithm efficiency. 360 361All of the algorithms presented in this text have a polynomial time work level. That is, of the form 362$O(n^k)$ for $n, k \in \Z^{+}$. This will help make useful comparisons in terms of the speed of the algorithms and how 363various optimizations will help pay off in the long run. 364 365\section{Exercises} 366Within the more advanced chapters a section will be set aside to give the reader some challenging exercises related to 367the discussion at hand. These exercises are not designed to be prize winning problems, but instead to be thought 368provoking. Wherever possible the problems are forward minded, stating problems that will be answered in subsequent 369chapters. The reader is encouraged to finish the exercises as they appear to get a better understanding of the 370subject material. 371 372That being said, the problems are designed to affirm knowledge of a particular subject matter. Students in particular 373are encouraged to verify they can answer the problems correctly before moving on. 374 375Similar to the exercises of \cite[pp. ix]{TAOCPV2} these exercises are given a scoring system based on the difficulty of 376the problem. However, unlike \cite{TAOCPV2} the problems do not get nearly as hard. The scoring of these 377exercises ranges from one (the easiest) to five (the hardest). The following table sumarizes the 378scoring system used. 379 380\begin{figure}[here] 381\begin{center} 382\begin{small} 383\begin{tabular}{|c|l|} 384\hline $\left [ 1 \right ]$ & An easy problem that should only take the reader a manner of \\ 385 & minutes to solve. Usually does not involve much computer time \\ 386 & to solve. \\ 387\hline $\left [ 2 \right ]$ & An easy problem that involves a marginal amount of computer \\ 388 & time usage. Usually requires a program to be written to \\ 389 & solve the problem. \\ 390\hline $\left [ 3 \right ]$ & A moderately hard problem that requires a non-trivial amount \\ 391 & of work. Usually involves trivial research and development of \\ 392 & new theory from the perspective of a student. \\ 393\hline $\left [ 4 \right ]$ & A moderately hard problem that involves a non-trivial amount \\ 394 & of work and research, the solution to which will demonstrate \\ 395 & a higher mastery of the subject matter. \\ 396\hline $\left [ 5 \right ]$ & A hard problem that involves concepts that are difficult for a \\ 397 & novice to solve. Solutions to these problems will demonstrate a \\ 398 & complete mastery of the given subject. \\ 399\hline 400\end{tabular} 401\end{small} 402\end{center} 403\caption{Exercise Scoring System} 404\end{figure} 405 406Problems at the first level are meant to be simple questions that the reader can answer quickly without programming a solution or 407devising new theory. These problems are quick tests to see if the material is understood. Problems at the second level 408are also designed to be easy but will require a program or algorithm to be implemented to arrive at the answer. These 409two levels are essentially entry level questions. 410 411Problems at the third level are meant to be a bit more difficult than the first two levels. The answer is often 412fairly obvious but arriving at an exacting solution requires some thought and skill. These problems will almost always 413involve devising a new algorithm or implementing a variation of another algorithm previously presented. Readers who can 414answer these questions will feel comfortable with the concepts behind the topic at hand. 415 416Problems at the fourth level are meant to be similar to those of the level three questions except they will require 417additional research to be completed. The reader will most likely not know the answer right away, nor will the text provide 418the exact details of the answer until a subsequent chapter. 419 420Problems at the fifth level are meant to be the hardest 421problems relative to all the other problems in the chapter. People who can correctly answer fifth level problems have a 422mastery of the subject matter at hand. 423 424Often problems will be tied together. The purpose of this is to start a chain of thought that will be discussed in future chapters. The reader 425is encouraged to answer the follow-up problems and try to draw the relevance of problems. 426 427\section{Introduction to LibTomMath} 428 429\subsection{What is LibTomMath?} 430LibTomMath is a free and open source multiple precision integer library written entirely in portable ISO C. By portable it 431is meant that the library does not contain any code that is computer platform dependent or otherwise problematic to use on 432any given platform. 433 434The library has been successfully tested under numerous operating systems including Unix\footnote{All of these 435trademarks belong to their respective rightful owners.}, MacOS, Windows, Linux, PalmOS and on standalone hardware such 436as the Gameboy Advance. The library is designed to contain enough functionality to be able to develop applications such 437as public key cryptosystems and still maintain a relatively small footprint. 438 439\subsection{Goals of LibTomMath} 440 441Libraries which obtain the most efficiency are rarely written in a high level programming language such as C. However, 442even though this library is written entirely in ISO C, considerable care has been taken to optimize the algorithm implementations within the 443library. Specifically the code has been written to work well with the GNU C Compiler (\textit{GCC}) on both x86 and ARM 444processors. Wherever possible, highly efficient algorithms, such as Karatsuba multiplication, sliding window 445exponentiation and Montgomery reduction have been provided to make the library more efficient. 446 447Even with the nearly optimal and specialized algorithms that have been included the Application Programing Interface 448(\textit{API}) has been kept as simple as possible. Often generic place holder routines will make use of specialized 449algorithms automatically without the developer's specific attention. One such example is the generic multiplication 450algorithm \textbf{mp\_mul()} which will automatically use Toom--Cook, Karatsuba, Comba or baseline multiplication 451based on the magnitude of the inputs and the configuration of the library. 452 453Making LibTomMath as efficient as possible is not the only goal of the LibTomMath project. Ideally the library should 454be source compatible with another popular library which makes it more attractive for developers to use. In this case the 455MPI library was used as a API template for all the basic functions. MPI was chosen because it is another library that fits 456in the same niche as LibTomMath. Even though LibTomMath uses MPI as the template for the function names and argument 457passing conventions, it has been written from scratch by Tom St Denis. 458 459The project is also meant to act as a learning tool for students, the logic being that no easy-to-follow ``bignum'' 460library exists which can be used to teach computer science students how to perform fast and reliable multiple precision 461integer arithmetic. To this end the source code has been given quite a few comments and algorithm discussion points. 462 463\section{Choice of LibTomMath} 464LibTomMath was chosen as the case study of this text not only because the author of both projects is one and the same but 465for more worthy reasons. Other libraries such as GMP \cite{GMP}, MPI \cite{MPI}, LIP \cite{LIP} and OpenSSL 466\cite{OPENSSL} have multiple precision integer arithmetic routines but would not be ideal for this text for 467reasons that will be explained in the following sub-sections. 468 469\subsection{Code Base} 470The LibTomMath code base is all portable ISO C source code. This means that there are no platform dependent conditional 471segments of code littered throughout the source. This clean and uncluttered approach to the library means that a 472developer can more readily discern the true intent of a given section of source code without trying to keep track of 473what conditional code will be used. 474 475The code base of LibTomMath is well organized. Each function is in its own separate source code file 476which allows the reader to find a given function very quickly. On average there are $76$ lines of code per source 477file which makes the source very easily to follow. By comparison MPI and LIP are single file projects making code tracing 478very hard. GMP has many conditional code segments which also hinder tracing. 479 480When compiled with GCC for the x86 processor and optimized for speed the entire library is approximately $100$KiB\footnote{The notation ``KiB'' means $2^{10}$ octets, similarly ``MiB'' means $2^{20}$ octets.} 481 which is fairly small compared to GMP (over $250$KiB). LibTomMath is slightly larger than MPI (which compiles to about 482$50$KiB) but LibTomMath is also much faster and more complete than MPI. 483 484\subsection{API Simplicity} 485LibTomMath is designed after the MPI library and shares the API design. Quite often programs that use MPI will build 486with LibTomMath without change. The function names correlate directly to the action they perform. Almost all of the 487functions share the same parameter passing convention. The learning curve is fairly shallow with the API provided 488which is an extremely valuable benefit for the student and developer alike. 489 490The LIP library is an example of a library with an API that is awkward to work with. LIP uses function names that are often ``compressed'' to 491illegible short hand. LibTomMath does not share this characteristic. 492 493The GMP library also does not return error codes. Instead it uses a POSIX.1 \cite{POSIX1} signal system where errors 494are signaled to the host application. This happens to be the fastest approach but definitely not the most versatile. In 495effect a math error (i.e. invalid input, heap error, etc) can cause a program to stop functioning which is definitely 496undersireable in many situations. 497 498\subsection{Optimizations} 499While LibTomMath is certainly not the fastest library (GMP often beats LibTomMath by a factor of two) it does 500feature a set of optimal algorithms for tasks such as modular reduction, exponentiation, multiplication and squaring. GMP 501and LIP also feature such optimizations while MPI only uses baseline algorithms with no optimizations. GMP lacks a few 502of the additional modular reduction optimizations that LibTomMath features\footnote{At the time of this writing GMP 503only had Barrett and Montgomery modular reduction algorithms.}. 504 505LibTomMath is almost always an order of magnitude faster than the MPI library at computationally expensive tasks such as modular 506exponentiation. In the grand scheme of ``bignum'' libraries LibTomMath is faster than the average library and usually 507slower than the best libraries such as GMP and OpenSSL by only a small factor. 508 509\subsection{Portability and Stability} 510LibTomMath will build ``out of the box'' on any platform equipped with a modern version of the GNU C Compiler 511(\textit{GCC}). This means that without changes the library will build without configuration or setting up any 512variables. LIP and MPI will build ``out of the box'' as well but have numerous known bugs. Most notably the author of 513MPI has recently stopped working on his library and LIP has long since been discontinued. 514 515GMP requires a configuration script to run and will not build out of the box. GMP and LibTomMath are still in active 516development and are very stable across a variety of platforms. 517 518\subsection{Choice} 519LibTomMath is a relatively compact, well documented, highly optimized and portable library which seems only natural for 520the case study of this text. Various source files from the LibTomMath project will be included within the text. However, 521the reader is encouraged to download their own copy of the library to actually be able to work with the library. 522 523\chapter{Getting Started} 524\section{Library Basics} 525The trick to writing any useful library of source code is to build a solid foundation and work outwards from it. First, 526a problem along with allowable solution parameters should be identified and analyzed. In this particular case the 527inability to accomodate multiple precision integers is the problem. Futhermore, the solution must be written 528as portable source code that is reasonably efficient across several different computer platforms. 529 530After a foundation is formed the remainder of the library can be designed and implemented in a hierarchical fashion. 531That is, to implement the lowest level dependencies first and work towards the most abstract functions last. For example, 532before implementing a modular exponentiation algorithm one would implement a modular reduction algorithm. 533By building outwards from a base foundation instead of using a parallel design methodology the resulting project is 534highly modular. Being highly modular is a desirable property of any project as it often means the resulting product 535has a small footprint and updates are easy to perform. 536 537Usually when I start a project I will begin with the header files. I define the data types I think I will need and 538prototype the initial functions that are not dependent on other functions (within the library). After I 539implement these base functions I prototype more dependent functions and implement them. The process repeats until 540I implement all of the functions I require. For example, in the case of LibTomMath I implemented functions such as 541mp\_init() well before I implemented mp\_mul() and even further before I implemented mp\_exptmod(). As an example as to 542why this design works note that the Karatsuba and Toom-Cook multipliers were written \textit{after} the 543dependent function mp\_exptmod() was written. Adding the new multiplication algorithms did not require changes to the 544mp\_exptmod() function itself and lowered the total cost of ownership (\textit{so to speak}) and of development 545for new algorithms. This methodology allows new algorithms to be tested in a complete framework with relative ease. 546 547FIGU,design_process,Design Flow of the First Few Original LibTomMath Functions. 548 549Only after the majority of the functions were in place did I pursue a less hierarchical approach to auditing and optimizing 550the source code. For example, one day I may audit the multipliers and the next day the polynomial basis functions. 551 552It only makes sense to begin the text with the preliminary data types and support algorithms required as well. 553This chapter discusses the core algorithms of the library which are the dependents for every other algorithm. 554 555\section{What is a Multiple Precision Integer?} 556Recall that most programming languages, in particular ISO C \cite{ISOC}, only have fixed precision data types that on their own cannot 557be used to represent values larger than their precision will allow. The purpose of multiple precision algorithms is 558to use fixed precision data types to create and manipulate multiple precision integers which may represent values 559that are very large. 560 561As a well known analogy, school children are taught how to form numbers larger than nine by prepending more radix ten digits. In the decimal system 562the largest single digit value is $9$. However, by concatenating digits together larger numbers may be represented. Newly prepended digits 563(\textit{to the left}) are said to be in a different power of ten column. That is, the number $123$ can be described as having a $1$ in the hundreds 564column, $2$ in the tens column and $3$ in the ones column. Or more formally $123 = 1 \cdot 10^2 + 2 \cdot 10^1 + 3 \cdot 10^0$. Computer based 565multiple precision arithmetic is essentially the same concept. Larger integers are represented by adjoining fixed 566precision computer words with the exception that a different radix is used. 567 568What most people probably do not think about explicitly are the various other attributes that describe a multiple precision 569integer. For example, the integer $154_{10}$ has two immediately obvious properties. First, the integer is positive, 570that is the sign of this particular integer is positive as opposed to negative. Second, the integer has three digits in 571its representation. There is an additional property that the integer posesses that does not concern pencil-and-paper 572arithmetic. The third property is how many digits placeholders are available to hold the integer. 573 574The human analogy of this third property is ensuring there is enough space on the paper to write the integer. For example, 575if one starts writing a large number too far to the right on a piece of paper they will have to erase it and move left. 576Similarly, computer algorithms must maintain strict control over memory usage to ensure that the digits of an integer 577will not exceed the allowed boundaries. These three properties make up what is known as a multiple precision 578integer or mp\_int for short. 579 580\subsection{The mp\_int Structure} 581\label{sec:MPINT} 582The mp\_int structure is the ISO C based manifestation of what represents a multiple precision integer. The ISO C standard does not provide for 583any such data type but it does provide for making composite data types known as structures. The following is the structure definition 584used within LibTomMath. 585 586\index{mp\_int} 587\begin{figure}[here] 588\begin{center} 589\begin{small} 590%\begin{verbatim} 591\begin{tabular}{|l|} 592\hline 593typedef struct \{ \\ 594\hspace{3mm}int used, alloc, sign;\\ 595\hspace{3mm}mp\_digit *dp;\\ 596\} \textbf{mp\_int}; \\ 597\hline 598\end{tabular} 599%\end{verbatim} 600\end{small} 601\caption{The mp\_int Structure} 602\label{fig:mpint} 603\end{center} 604\end{figure} 605 606The mp\_int structure (fig. \ref{fig:mpint}) can be broken down as follows. 607 608\begin{enumerate} 609\item The \textbf{used} parameter denotes how many digits of the array \textbf{dp} contain the digits used to represent 610a given integer. The \textbf{used} count must be positive (or zero) and may not exceed the \textbf{alloc} count. 611 612\item The \textbf{alloc} parameter denotes how 613many digits are available in the array to use by functions before it has to increase in size. When the \textbf{used} count 614of a result would exceed the \textbf{alloc} count all of the algorithms will automatically increase the size of the 615array to accommodate the precision of the result. 616 617\item The pointer \textbf{dp} points to a dynamically allocated array of digits that represent the given multiple 618precision integer. It is padded with $(\textbf{alloc} - \textbf{used})$ zero digits. The array is maintained in a least 619significant digit order. As a pencil and paper analogy the array is organized such that the right most digits are stored 620first starting at the location indexed by zero\footnote{In C all arrays begin at zero.} in the array. For example, 621if \textbf{dp} contains $\lbrace a, b, c, \ldots \rbrace$ where \textbf{dp}$_0 = a$, \textbf{dp}$_1 = b$, \textbf{dp}$_2 = c$, $\ldots$ then 622it would represent the integer $a + b\beta + c\beta^2 + \ldots$ 623 624\index{MP\_ZPOS} \index{MP\_NEG} 625\item The \textbf{sign} parameter denotes the sign as either zero/positive (\textbf{MP\_ZPOS}) or negative (\textbf{MP\_NEG}). 626\end{enumerate} 627 628\subsubsection{Valid mp\_int Structures} 629Several rules are placed on the state of an mp\_int structure and are assumed to be followed for reasons of efficiency. 630The only exceptions are when the structure is passed to initialization functions such as mp\_init() and mp\_init\_copy(). 631 632\begin{enumerate} 633\item The value of \textbf{alloc} may not be less than one. That is \textbf{dp} always points to a previously allocated 634array of digits. 635\item The value of \textbf{used} may not exceed \textbf{alloc} and must be greater than or equal to zero. 636\item The value of \textbf{used} implies the digit at index $(used - 1)$ of the \textbf{dp} array is non-zero. That is, 637leading zero digits in the most significant positions must be trimmed. 638 \begin{enumerate} 639 \item Digits in the \textbf{dp} array at and above the \textbf{used} location must be zero. 640 \end{enumerate} 641\item The value of \textbf{sign} must be \textbf{MP\_ZPOS} if \textbf{used} is zero; 642this represents the mp\_int value of zero. 643\end{enumerate} 644 645\section{Argument Passing} 646A convention of argument passing must be adopted early on in the development of any library. Making the function 647prototypes consistent will help eliminate many headaches in the future as the library grows to significant complexity. 648In LibTomMath the multiple precision integer functions accept parameters from left to right as pointers to mp\_int 649structures. That means that the source (input) operands are placed on the left and the destination (output) on the right. 650Consider the following examples. 651 652\begin{verbatim} 653 mp_mul(&a, &b, &c); /* c = a * b */ 654 mp_add(&a, &b, &a); /* a = a + b */ 655 mp_sqr(&a, &b); /* b = a * a */ 656\end{verbatim} 657 658The left to right order is a fairly natural way to implement the functions since it lets the developer read aloud the 659functions and make sense of them. For example, the first function would read ``multiply a and b and store in c''. 660 661Certain libraries (\textit{LIP by Lenstra for instance}) accept parameters the other way around, to mimic the order 662of assignment expressions. That is, the destination (output) is on the left and arguments (inputs) are on the right. In 663truth, it is entirely a matter of preference. In the case of LibTomMath the convention from the MPI library has been 664adopted. 665 666Another very useful design consideration, provided for in LibTomMath, is whether to allow argument sources to also be a 667destination. For example, the second example (\textit{mp\_add}) adds $a$ to $b$ and stores in $a$. This is an important 668feature to implement since it allows the calling functions to cut down on the number of variables it must maintain. 669However, to implement this feature specific care has to be given to ensure the destination is not modified before the 670source is fully read. 671 672\section{Return Values} 673A well implemented application, no matter what its purpose, should trap as many runtime errors as possible and return them 674to the caller. By catching runtime errors a library can be guaranteed to prevent undefined behaviour. However, the end 675developer can still manage to cause a library to crash. For example, by passing an invalid pointer an application may 676fault by dereferencing memory not owned by the application. 677 678In the case of LibTomMath the only errors that are checked for are related to inappropriate inputs (division by zero for 679instance) and memory allocation errors. It will not check that the mp\_int passed to any function is valid nor 680will it check pointers for validity. Any function that can cause a runtime error will return an error code as an 681\textbf{int} data type with one of the following values (fig \ref{fig:errcodes}). 682 683\index{MP\_OKAY} \index{MP\_VAL} \index{MP\_MEM} 684\begin{figure}[here] 685\begin{center} 686\begin{tabular}{|l|l|} 687\hline \textbf{Value} & \textbf{Meaning} \\ 688\hline \textbf{MP\_OKAY} & The function was successful \\ 689\hline \textbf{MP\_VAL} & One of the input value(s) was invalid \\ 690\hline \textbf{MP\_MEM} & The function ran out of heap memory \\ 691\hline 692\end{tabular} 693\end{center} 694\caption{LibTomMath Error Codes} 695\label{fig:errcodes} 696\end{figure} 697 698When an error is detected within a function it should free any memory it allocated, often during the initialization of 699temporary mp\_ints, and return as soon as possible. The goal is to leave the system in the same state it was when the 700function was called. Error checking with this style of API is fairly simple. 701 702\begin{verbatim} 703 int err; 704 if ((err = mp_add(&a, &b, &c)) != MP_OKAY) { 705 printf("Error: %s\n", mp_error_to_string(err)); 706 exit(EXIT_FAILURE); 707 } 708\end{verbatim} 709 710The GMP \cite{GMP} library uses C style \textit{signals} to flag errors which is of questionable use. Not all errors are fatal 711and it was not deemed ideal by the author of LibTomMath to force developers to have signal handlers for such cases. 712 713\section{Initialization and Clearing} 714The logical starting point when actually writing multiple precision integer functions is the initialization and 715clearing of the mp\_int structures. These two algorithms will be used by the majority of the higher level algorithms. 716 717Given the basic mp\_int structure an initialization routine must first allocate memory to hold the digits of 718the integer. Often it is optimal to allocate a sufficiently large pre-set number of digits even though 719the initial integer will represent zero. If only a single digit were allocated quite a few subsequent re-allocations 720would occur when operations are performed on the integers. There is a tradeoff between how many default digits to allocate 721and how many re-allocations are tolerable. Obviously allocating an excessive amount of digits initially will waste 722memory and become unmanageable. 723 724If the memory for the digits has been successfully allocated then the rest of the members of the structure must 725be initialized. Since the initial state of an mp\_int is to represent the zero integer, the allocated digits must be set 726to zero. The \textbf{used} count set to zero and \textbf{sign} set to \textbf{MP\_ZPOS}. 727 728\subsection{Initializing an mp\_int} 729An mp\_int is said to be initialized if it is set to a valid, preferably default, state such that all of the members of the 730structure are set to valid values. The mp\_init algorithm will perform such an action. 731 732\index{mp\_init} 733\begin{figure}[here] 734\begin{center} 735\begin{tabular}{l} 736\hline Algorithm \textbf{mp\_init}. \\ 737\textbf{Input}. An mp\_int $a$ \\ 738\textbf{Output}. Allocate memory and initialize $a$ to a known valid mp\_int state. \\ 739\hline \\ 7401. Allocate memory for \textbf{MP\_PREC} digits. \\ 7412. If the allocation failed return(\textit{MP\_MEM}) \\ 7423. for $n$ from $0$ to $MP\_PREC - 1$ do \\ 743\hspace{3mm}3.1 $a_n \leftarrow 0$\\ 7444. $a.sign \leftarrow MP\_ZPOS$\\ 7455. $a.used \leftarrow 0$\\ 7466. $a.alloc \leftarrow MP\_PREC$\\ 7477. Return(\textit{MP\_OKAY})\\ 748\hline 749\end{tabular} 750\end{center} 751\caption{Algorithm mp\_init} 752\end{figure} 753 754\textbf{Algorithm mp\_init.} 755The purpose of this function is to initialize an mp\_int structure so that the rest of the library can properly 756manipulte it. It is assumed that the input may not have had any of its members previously initialized which is certainly 757a valid assumption if the input resides on the stack. 758 759Before any of the members such as \textbf{sign}, \textbf{used} or \textbf{alloc} are initialized the memory for 760the digits is allocated. If this fails the function returns before setting any of the other members. The \textbf{MP\_PREC} 761name represents a constant\footnote{Defined in the ``tommath.h'' header file within LibTomMath.} 762used to dictate the minimum precision of newly initialized mp\_int integers. Ideally, it is at least equal to the smallest 763precision number you'll be working with. 764 765Allocating a block of digits at first instead of a single digit has the benefit of lowering the number of usually slow 766heap operations later functions will have to perform in the future. If \textbf{MP\_PREC} is set correctly the slack 767memory and the number of heap operations will be trivial. 768 769Once the allocation has been made the digits have to be set to zero as well as the \textbf{used}, \textbf{sign} and 770\textbf{alloc} members initialized. This ensures that the mp\_int will always represent the default state of zero regardless 771of the original condition of the input. 772 773\textbf{Remark.} 774This function introduces the idiosyncrasy that all iterative loops, commonly initiated with the ``for'' keyword, iterate incrementally 775when the ``to'' keyword is placed between two expressions. For example, ``for $a$ from $b$ to $c$ do'' means that 776a subsequent expression (or body of expressions) are to be evaluated upto $c - b$ times so long as $b \le c$. In each 777iteration the variable $a$ is substituted for a new integer that lies inclusively between $b$ and $c$. If $b > c$ occured 778the loop would not iterate. By contrast if the ``downto'' keyword were used in place of ``to'' the loop would iterate 779decrementally. 780 781EXAM,bn_mp_init.c 782 783One immediate observation of this initializtion function is that it does not return a pointer to a mp\_int structure. It 784is assumed that the caller has already allocated memory for the mp\_int structure, typically on the application stack. The 785call to mp\_init() is used only to initialize the members of the structure to a known default state. 786 787Here we see (line @23,XMALLOC@) the memory allocation is performed first. This allows us to exit cleanly and quickly 788if there is an error. If the allocation fails the routine will return \textbf{MP\_MEM} to the caller to indicate there 789was a memory error. The function XMALLOC is what actually allocates the memory. Technically XMALLOC is not a function 790but a macro defined in ``tommath.h``. By default, XMALLOC will evaluate to malloc() which is the C library's built--in 791memory allocation routine. 792 793In order to assure the mp\_int is in a known state the digits must be set to zero. On most platforms this could have been 794accomplished by using calloc() instead of malloc(). However, to correctly initialize a integer type to a given value in a 795portable fashion you have to actually assign the value. The for loop (line @28,for@) performs this required 796operation. 797 798After the memory has been successfully initialized the remainder of the members are initialized 799(lines @29,used@ through @31,sign@) to their respective default states. At this point the algorithm has succeeded and 800a success code is returned to the calling function. If this function returns \textbf{MP\_OKAY} it is safe to assume the 801mp\_int structure has been properly initialized and is safe to use with other functions within the library. 802 803\subsection{Clearing an mp\_int} 804When an mp\_int is no longer required by the application, the memory that has been allocated for its digits must be 805returned to the application's memory pool with the mp\_clear algorithm. 806 807\begin{figure}[here] 808\begin{center} 809\begin{tabular}{l} 810\hline Algorithm \textbf{mp\_clear}. \\ 811\textbf{Input}. An mp\_int $a$ \\ 812\textbf{Output}. The memory for $a$ shall be deallocated. \\ 813\hline \\ 8141. If $a$ has been previously freed then return(\textit{MP\_OKAY}). \\ 8152. for $n$ from 0 to $a.used - 1$ do \\ 816\hspace{3mm}2.1 $a_n \leftarrow 0$ \\ 8173. Free the memory allocated for the digits of $a$. \\ 8184. $a.used \leftarrow 0$ \\ 8195. $a.alloc \leftarrow 0$ \\ 8206. $a.sign \leftarrow MP\_ZPOS$ \\ 8217. Return(\textit{MP\_OKAY}). \\ 822\hline 823\end{tabular} 824\end{center} 825\caption{Algorithm mp\_clear} 826\end{figure} 827 828\textbf{Algorithm mp\_clear.} 829This algorithm accomplishes two goals. First, it clears the digits and the other mp\_int members. This ensures that 830if a developer accidentally re-uses a cleared structure it is less likely to cause problems. The second goal 831is to free the allocated memory. 832 833The logic behind the algorithm is extended by marking cleared mp\_int structures so that subsequent calls to this 834algorithm will not try to free the memory multiple times. Cleared mp\_ints are detectable by having a pre-defined invalid 835digit pointer \textbf{dp} setting. 836 837Once an mp\_int has been cleared the mp\_int structure is no longer in a valid state for any other algorithm 838with the exception of algorithms mp\_init, mp\_init\_copy, mp\_init\_size and mp\_clear. 839 840EXAM,bn_mp_clear.c 841 842The algorithm only operates on the mp\_int if it hasn't been previously cleared. The if statement (line @23,a->dp != NULL@) 843checks to see if the \textbf{dp} member is not \textbf{NULL}. If the mp\_int is a valid mp\_int then \textbf{dp} cannot be 844\textbf{NULL} in which case the if statement will evaluate to true. 845 846The digits of the mp\_int are cleared by the for loop (line @25,for@) which assigns a zero to every digit. Similar to mp\_init() 847the digits are assigned zero instead of using block memory operations (such as memset()) since this is more portable. 848 849The digits are deallocated off the heap via the XFREE macro. Similar to XMALLOC the XFREE macro actually evaluates to 850a standard C library function. In this case the free() function. Since free() only deallocates the memory the pointer 851still has to be reset to \textbf{NULL} manually (line @33,NULL@). 852 853Now that the digits have been cleared and deallocated the other members are set to their final values (lines @34,= 0@ and @35,ZPOS@). 854 855\section{Maintenance Algorithms} 856 857The previous sections describes how to initialize and clear an mp\_int structure. To further support operations 858that are to be performed on mp\_int structures (such as addition and multiplication) the dependent algorithms must be 859able to augment the precision of an mp\_int and 860initialize mp\_ints with differing initial conditions. 861 862These algorithms complete the set of low level algorithms required to work with mp\_int structures in the higher level 863algorithms such as addition, multiplication and modular exponentiation. 864 865\subsection{Augmenting an mp\_int's Precision} 866When storing a value in an mp\_int structure, a sufficient number of digits must be available to accomodate the entire 867result of an operation without loss of precision. Quite often the size of the array given by the \textbf{alloc} member 868is large enough to simply increase the \textbf{used} digit count. However, when the size of the array is too small it 869must be re-sized appropriately to accomodate the result. The mp\_grow algorithm will provide this functionality. 870 871\newpage\begin{figure}[here] 872\begin{center} 873\begin{tabular}{l} 874\hline Algorithm \textbf{mp\_grow}. \\ 875\textbf{Input}. An mp\_int $a$ and an integer $b$. \\ 876\textbf{Output}. $a$ is expanded to accomodate $b$ digits. \\ 877\hline \\ 8781. if $a.alloc \ge b$ then return(\textit{MP\_OKAY}) \\ 8792. $u \leftarrow b\mbox{ (mod }MP\_PREC\mbox{)}$ \\ 8803. $v \leftarrow b + 2 \cdot MP\_PREC - u$ \\ 8814. Re-allocate the array of digits $a$ to size $v$ \\ 8825. If the allocation failed then return(\textit{MP\_MEM}). \\ 8836. for n from a.alloc to $v - 1$ do \\ 884\hspace{+3mm}6.1 $a_n \leftarrow 0$ \\ 8857. $a.alloc \leftarrow v$ \\ 8868. Return(\textit{MP\_OKAY}) \\ 887\hline 888\end{tabular} 889\end{center} 890\caption{Algorithm mp\_grow} 891\end{figure} 892 893\textbf{Algorithm mp\_grow.} 894It is ideal to prevent re-allocations from being performed if they are not required (step one). This is useful to 895prevent mp\_ints from growing excessively in code that erroneously calls mp\_grow. 896 897The requested digit count is padded up to next multiple of \textbf{MP\_PREC} plus an additional \textbf{MP\_PREC} (steps two and three). 898This helps prevent many trivial reallocations that would grow an mp\_int by trivially small values. 899 900It is assumed that the reallocation (step four) leaves the lower $a.alloc$ digits of the mp\_int intact. This is much 901akin to how the \textit{realloc} function from the standard C library works. Since the newly allocated digits are 902assumed to contain undefined values they are initially set to zero. 903 904EXAM,bn_mp_grow.c 905 906A quick optimization is to first determine if a memory re-allocation is required at all. The if statement (line @24,alloc@) checks 907if the \textbf{alloc} member of the mp\_int is smaller than the requested digit count. If the count is not larger than \textbf{alloc} 908the function skips the re-allocation part thus saving time. 909 910When a re-allocation is performed it is turned into an optimal request to save time in the future. The requested digit count is 911padded upwards to 2nd multiple of \textbf{MP\_PREC} larger than \textbf{alloc} (line @25, size@). The XREALLOC function is used 912to re-allocate the memory. As per the other functions XREALLOC is actually a macro which evaluates to realloc by default. The realloc 913function leaves the base of the allocation intact which means the first \textbf{alloc} digits of the mp\_int are the same as before 914the re-allocation. All that is left is to clear the newly allocated digits and return. 915 916Note that the re-allocation result is actually stored in a temporary pointer $tmp$. This is to allow this function to return 917an error with a valid pointer. Earlier releases of the library stored the result of XREALLOC into the mp\_int $a$. That would 918result in a memory leak if XREALLOC ever failed. 919 920\subsection{Initializing Variable Precision mp\_ints} 921Occasionally the number of digits required will be known in advance of an initialization, based on, for example, the size 922of input mp\_ints to a given algorithm. The purpose of algorithm mp\_init\_size is similar to mp\_init except that it 923will allocate \textit{at least} a specified number of digits. 924 925\begin{figure}[here] 926\begin{small} 927\begin{center} 928\begin{tabular}{l} 929\hline Algorithm \textbf{mp\_init\_size}. \\ 930\textbf{Input}. An mp\_int $a$ and the requested number of digits $b$. \\ 931\textbf{Output}. $a$ is initialized to hold at least $b$ digits. \\ 932\hline \\ 9331. $u \leftarrow b \mbox{ (mod }MP\_PREC\mbox{)}$ \\ 9342. $v \leftarrow b + 2 \cdot MP\_PREC - u$ \\ 9353. Allocate $v$ digits. \\ 9364. for $n$ from $0$ to $v - 1$ do \\ 937\hspace{3mm}4.1 $a_n \leftarrow 0$ \\ 9385. $a.sign \leftarrow MP\_ZPOS$\\ 9396. $a.used \leftarrow 0$\\ 9407. $a.alloc \leftarrow v$\\ 9418. Return(\textit{MP\_OKAY})\\ 942\hline 943\end{tabular} 944\end{center} 945\end{small} 946\caption{Algorithm mp\_init\_size} 947\end{figure} 948 949\textbf{Algorithm mp\_init\_size.} 950This algorithm will initialize an mp\_int structure $a$ like algorithm mp\_init with the exception that the number of 951digits allocated can be controlled by the second input argument $b$. The input size is padded upwards so it is a 952multiple of \textbf{MP\_PREC} plus an additional \textbf{MP\_PREC} digits. This padding is used to prevent trivial 953allocations from becoming a bottleneck in the rest of the algorithms. 954 955Like algorithm mp\_init, the mp\_int structure is initialized to a default state representing the integer zero. This 956particular algorithm is useful if it is known ahead of time the approximate size of the input. If the approximation is 957correct no further memory re-allocations are required to work with the mp\_int. 958 959EXAM,bn_mp_init_size.c 960 961The number of digits $b$ requested is padded (line @22,MP_PREC@) by first augmenting it to the next multiple of 962\textbf{MP\_PREC} and then adding \textbf{MP\_PREC} to the result. If the memory can be successfully allocated the 963mp\_int is placed in a default state representing the integer zero. Otherwise, the error code \textbf{MP\_MEM} will be 964returned (line @27,return@). 965 966The digits are allocated and set to zero at the same time with the calloc() function (line @25,XCALLOC@). The 967\textbf{used} count is set to zero, the \textbf{alloc} count set to the padded digit count and the \textbf{sign} flag set 968to \textbf{MP\_ZPOS} to achieve a default valid mp\_int state (lines @29,used@, @30,alloc@ and @31,sign@). If the function 969returns succesfully then it is correct to assume that the mp\_int structure is in a valid state for the remainder of the 970functions to work with. 971 972\subsection{Multiple Integer Initializations and Clearings} 973Occasionally a function will require a series of mp\_int data types to be made available simultaneously. 974The purpose of algorithm mp\_init\_multi is to initialize a variable length array of mp\_int structures in a single 975statement. It is essentially a shortcut to multiple initializations. 976 977\newpage\begin{figure}[here] 978\begin{center} 979\begin{tabular}{l} 980\hline Algorithm \textbf{mp\_init\_multi}. \\ 981\textbf{Input}. Variable length array $V_k$ of mp\_int variables of length $k$. \\ 982\textbf{Output}. The array is initialized such that each mp\_int of $V_k$ is ready to use. \\ 983\hline \\ 9841. for $n$ from 0 to $k - 1$ do \\ 985\hspace{+3mm}1.1. Initialize the mp\_int $V_n$ (\textit{mp\_init}) \\ 986\hspace{+3mm}1.2. If initialization failed then do \\ 987\hspace{+6mm}1.2.1. for $j$ from $0$ to $n$ do \\ 988\hspace{+9mm}1.2.1.1. Free the mp\_int $V_j$ (\textit{mp\_clear}) \\ 989\hspace{+6mm}1.2.2. Return(\textit{MP\_MEM}) \\ 9902. Return(\textit{MP\_OKAY}) \\ 991\hline 992\end{tabular} 993\end{center} 994\caption{Algorithm mp\_init\_multi} 995\end{figure} 996 997\textbf{Algorithm mp\_init\_multi.} 998The algorithm will initialize the array of mp\_int variables one at a time. If a runtime error has been detected 999(\textit{step 1.2}) all of the previously initialized variables are cleared. The goal is an ``all or nothing'' 1000initialization which allows for quick recovery from runtime errors. 1001 1002EXAM,bn_mp_init_multi.c 1003 1004This function intializes a variable length list of mp\_int structure pointers. However, instead of having the mp\_int 1005structures in an actual C array they are simply passed as arguments to the function. This function makes use of the 1006``...'' argument syntax of the C programming language. The list is terminated with a final \textbf{NULL} argument 1007appended on the right. 1008 1009The function uses the ``stdarg.h'' \textit{va} functions to step portably through the arguments to the function. A count 1010$n$ of succesfully initialized mp\_int structures is maintained (line @47,n++@) such that if a failure does occur, 1011the algorithm can backtrack and free the previously initialized structures (lines @27,if@ to @46,}@). 1012 1013 1014\subsection{Clamping Excess Digits} 1015When a function anticipates a result will be $n$ digits it is simpler to assume this is true within the body of 1016the function instead of checking during the computation. For example, a multiplication of a $i$ digit number by a 1017$j$ digit produces a result of at most $i + j$ digits. It is entirely possible that the result is $i + j - 1$ 1018though, with no final carry into the last position. However, suppose the destination had to be first expanded 1019(\textit{via mp\_grow}) to accomodate $i + j - 1$ digits than further expanded to accomodate the final carry. 1020That would be a considerable waste of time since heap operations are relatively slow. 1021 1022The ideal solution is to always assume the result is $i + j$ and fix up the \textbf{used} count after the function 1023terminates. This way a single heap operation (\textit{at most}) is required. However, if the result was not checked 1024there would be an excess high order zero digit. 1025 1026For example, suppose the product of two integers was $x_n = (0x_{n-1}x_{n-2}...x_0)_{\beta}$. The leading zero digit 1027will not contribute to the precision of the result. In fact, through subsequent operations more leading zero digits would 1028accumulate to the point the size of the integer would be prohibitive. As a result even though the precision is very 1029low the representation is excessively large. 1030 1031The mp\_clamp algorithm is designed to solve this very problem. It will trim high-order zeros by decrementing the 1032\textbf{used} count until a non-zero most significant digit is found. Also in this system, zero is considered to be a 1033positive number which means that if the \textbf{used} count is decremented to zero, the sign must be set to 1034\textbf{MP\_ZPOS}. 1035 1036\begin{figure}[here] 1037\begin{center} 1038\begin{tabular}{l} 1039\hline Algorithm \textbf{mp\_clamp}. \\ 1040\textbf{Input}. An mp\_int $a$ \\ 1041\textbf{Output}. Any excess leading zero digits of $a$ are removed \\ 1042\hline \\ 10431. while $a.used > 0$ and $a_{a.used - 1} = 0$ do \\ 1044\hspace{+3mm}1.1 $a.used \leftarrow a.used - 1$ \\ 10452. if $a.used = 0$ then do \\ 1046\hspace{+3mm}2.1 $a.sign \leftarrow MP\_ZPOS$ \\ 1047\hline \\ 1048\end{tabular} 1049\end{center} 1050\caption{Algorithm mp\_clamp} 1051\end{figure} 1052 1053\textbf{Algorithm mp\_clamp.} 1054As can be expected this algorithm is very simple. The loop on step one is expected to iterate only once or twice at 1055the most. For example, this will happen in cases where there is not a carry to fill the last position. Step two fixes the sign for 1056when all of the digits are zero to ensure that the mp\_int is valid at all times. 1057 1058EXAM,bn_mp_clamp.c 1059 1060Note on line @27,while@ how to test for the \textbf{used} count is made on the left of the \&\& operator. In the C programming 1061language the terms to \&\& are evaluated left to right with a boolean short-circuit if any condition fails. This is 1062important since if the \textbf{used} is zero the test on the right would fetch below the array. That is obviously 1063undesirable. The parenthesis on line @28,a->used@ is used to make sure the \textbf{used} count is decremented and not 1064the pointer ``a''. 1065 1066\section*{Exercises} 1067\begin{tabular}{cl} 1068$\left [ 1 \right ]$ & Discuss the relevance of the \textbf{used} member of the mp\_int structure. \\ 1069 & \\ 1070$\left [ 1 \right ]$ & Discuss the consequences of not using padding when performing allocations. \\ 1071 & \\ 1072$\left [ 2 \right ]$ & Estimate an ideal value for \textbf{MP\_PREC} when performing 1024-bit RSA \\ 1073 & encryption when $\beta = 2^{28}$. \\ 1074 & \\ 1075$\left [ 1 \right ]$ & Discuss the relevance of the algorithm mp\_clamp. What does it prevent? \\ 1076 & \\ 1077$\left [ 1 \right ]$ & Give an example of when the algorithm mp\_init\_copy might be useful. \\ 1078 & \\ 1079\end{tabular} 1080 1081 1082%%% 1083% CHAPTER FOUR 1084%%% 1085 1086\chapter{Basic Operations} 1087 1088\section{Introduction} 1089In the previous chapter a series of low level algorithms were established that dealt with initializing and maintaining 1090mp\_int structures. This chapter will discuss another set of seemingly non-algebraic algorithms which will form the low 1091level basis of the entire library. While these algorithm are relatively trivial it is important to understand how they 1092work before proceeding since these algorithms will be used almost intrinsically in the following chapters. 1093 1094The algorithms in this chapter deal primarily with more ``programmer'' related tasks such as creating copies of 1095mp\_int structures, assigning small values to mp\_int structures and comparisons of the values mp\_int structures 1096represent. 1097 1098\section{Assigning Values to mp\_int Structures} 1099\subsection{Copying an mp\_int} 1100Assigning the value that a given mp\_int structure represents to another mp\_int structure shall be known as making 1101a copy for the purposes of this text. The copy of the mp\_int will be a separate entity that represents the same 1102value as the mp\_int it was copied from. The mp\_copy algorithm provides this functionality. 1103 1104\newpage\begin{figure}[here] 1105\begin{center} 1106\begin{tabular}{l} 1107\hline Algorithm \textbf{mp\_copy}. \\ 1108\textbf{Input}. An mp\_int $a$ and $b$. \\ 1109\textbf{Output}. Store a copy of $a$ in $b$. \\ 1110\hline \\ 11111. If $b.alloc < a.used$ then grow $b$ to $a.used$ digits. (\textit{mp\_grow}) \\ 11122. for $n$ from 0 to $a.used - 1$ do \\ 1113\hspace{3mm}2.1 $b_{n} \leftarrow a_{n}$ \\ 11143. for $n$ from $a.used$ to $b.used - 1$ do \\ 1115\hspace{3mm}3.1 $b_{n} \leftarrow 0$ \\ 11164. $b.used \leftarrow a.used$ \\ 11175. $b.sign \leftarrow a.sign$ \\ 11186. return(\textit{MP\_OKAY}) \\ 1119\hline 1120\end{tabular} 1121\end{center} 1122\caption{Algorithm mp\_copy} 1123\end{figure} 1124 1125\textbf{Algorithm mp\_copy.} 1126This algorithm copies the mp\_int $a$ such that upon succesful termination of the algorithm the mp\_int $b$ will 1127represent the same integer as the mp\_int $a$. The mp\_int $b$ shall be a complete and distinct copy of the 1128mp\_int $a$ meaing that the mp\_int $a$ can be modified and it shall not affect the value of the mp\_int $b$. 1129 1130If $b$ does not have enough room for the digits of $a$ it must first have its precision augmented via the mp\_grow 1131algorithm. The digits of $a$ are copied over the digits of $b$ and any excess digits of $b$ are set to zero (step two 1132and three). The \textbf{used} and \textbf{sign} members of $a$ are finally copied over the respective members of 1133$b$. 1134 1135\textbf{Remark.} This algorithm also introduces a new idiosyncrasy that will be used throughout the rest of the 1136text. The error return codes of other algorithms are not explicitly checked in the pseudo-code presented. For example, in 1137step one of the mp\_copy algorithm the return of mp\_grow is not explicitly checked to ensure it succeeded. Text space is 1138limited so it is assumed that if a algorithm fails it will clear all temporarily allocated mp\_ints and return 1139the error code itself. However, the C code presented will demonstrate all of the error handling logic required to 1140implement the pseudo-code. 1141 1142EXAM,bn_mp_copy.c 1143 1144Occasionally a dependent algorithm may copy an mp\_int effectively into itself such as when the input and output 1145mp\_int structures passed to a function are one and the same. For this case it is optimal to return immediately without 1146copying digits (line @24,a == b@). 1147 1148The mp\_int $b$ must have enough digits to accomodate the used digits of the mp\_int $a$. If $b.alloc$ is less than 1149$a.used$ the algorithm mp\_grow is used to augment the precision of $b$ (lines @29,alloc@ to @33,}@). In order to 1150simplify the inner loop that copies the digits from $a$ to $b$, two aliases $tmpa$ and $tmpb$ point directly at the digits 1151of the mp\_ints $a$ and $b$ respectively. These aliases (lines @42,tmpa@ and @45,tmpb@) allow the compiler to access the digits without first dereferencing the 1152mp\_int pointers and then subsequently the pointer to the digits. 1153 1154After the aliases are established the digits from $a$ are copied into $b$ (lines @48,for@ to @50,}@) and then the excess 1155digits of $b$ are set to zero (lines @53,for@ to @55,}@). Both ``for'' loops make use of the pointer aliases and in 1156fact the alias for $b$ is carried through into the second ``for'' loop to clear the excess digits. This optimization 1157allows the alias to stay in a machine register fairly easy between the two loops. 1158 1159\textbf{Remarks.} The use of pointer aliases is an implementation methodology first introduced in this function that will 1160be used considerably in other functions. Technically, a pointer alias is simply a short hand alias used to lower the 1161number of pointer dereferencing operations required to access data. For example, a for loop may resemble 1162 1163\begin{alltt} 1164for (x = 0; x < 100; x++) \{ 1165 a->num[4]->dp[x] = 0; 1166\} 1167\end{alltt} 1168 1169This could be re-written using aliases as 1170 1171\begin{alltt} 1172mp_digit *tmpa; 1173a = a->num[4]->dp; 1174for (x = 0; x < 100; x++) \{ 1175 *a++ = 0; 1176\} 1177\end{alltt} 1178 1179In this case an alias is used to access the 1180array of digits within an mp\_int structure directly. It may seem that a pointer alias is strictly not required 1181as a compiler may optimize out the redundant pointer operations. However, there are two dominant reasons to use aliases. 1182 1183The first reason is that most compilers will not effectively optimize pointer arithmetic. For example, some optimizations 1184may work for the Microsoft Visual C++ compiler (MSVC) and not for the GNU C Compiler (GCC). Also some optimizations may 1185work for GCC and not MSVC. As such it is ideal to find a common ground for as many compilers as possible. Pointer 1186aliases optimize the code considerably before the compiler even reads the source code which means the end compiled code 1187stands a better chance of being faster. 1188 1189The second reason is that pointer aliases often can make an algorithm simpler to read. Consider the first ``for'' 1190loop of the function mp\_copy() re-written to not use pointer aliases. 1191 1192\begin{alltt} 1193 /* copy all the digits */ 1194 for (n = 0; n < a->used; n++) \{ 1195 b->dp[n] = a->dp[n]; 1196 \} 1197\end{alltt} 1198 1199Whether this code is harder to read depends strongly on the individual. However, it is quantifiably slightly more 1200complicated as there are four variables within the statement instead of just two. 1201 1202\subsubsection{Nested Statements} 1203Another commonly used technique in the source routines is that certain sections of code are nested. This is used in 1204particular with the pointer aliases to highlight code phases. For example, a Comba multiplier (discussed in chapter six) 1205will typically have three different phases. First the temporaries are initialized, then the columns calculated and 1206finally the carries are propagated. In this example the middle column production phase will typically be nested as it 1207uses temporary variables and aliases the most. 1208 1209The nesting also simplies the source code as variables that are nested are only valid for their scope. As a result 1210the various temporary variables required do not propagate into other sections of code. 1211 1212 1213\subsection{Creating a Clone} 1214Another common operation is to make a local temporary copy of an mp\_int argument. To initialize an mp\_int 1215and then copy another existing mp\_int into the newly intialized mp\_int will be known as creating a clone. This is 1216useful within functions that need to modify an argument but do not wish to actually modify the original copy. The 1217mp\_init\_copy algorithm has been designed to help perform this task. 1218 1219\begin{figure}[here] 1220\begin{center} 1221\begin{tabular}{l} 1222\hline Algorithm \textbf{mp\_init\_copy}. \\ 1223\textbf{Input}. An mp\_int $a$ and $b$\\ 1224\textbf{Output}. $a$ is initialized to be a copy of $b$. \\ 1225\hline \\ 12261. Init $a$. (\textit{mp\_init}) \\ 12272. Copy $b$ to $a$. (\textit{mp\_copy}) \\ 12283. Return the status of the copy operation. \\ 1229\hline 1230\end{tabular} 1231\end{center} 1232\caption{Algorithm mp\_init\_copy} 1233\end{figure} 1234 1235\textbf{Algorithm mp\_init\_copy.} 1236This algorithm will initialize an mp\_int variable and copy another previously initialized mp\_int variable into it. As 1237such this algorithm will perform two operations in one step. 1238 1239EXAM,bn_mp_init_copy.c 1240 1241This will initialize \textbf{a} and make it a verbatim copy of the contents of \textbf{b}. Note that 1242\textbf{a} will have its own memory allocated which means that \textbf{b} may be cleared after the call 1243and \textbf{a} will be left intact. 1244 1245\section{Zeroing an Integer} 1246Reseting an mp\_int to the default state is a common step in many algorithms. The mp\_zero algorithm will be the algorithm used to 1247perform this task. 1248 1249\begin{figure}[here] 1250\begin{center} 1251\begin{tabular}{l} 1252\hline Algorithm \textbf{mp\_zero}. \\ 1253\textbf{Input}. An mp\_int $a$ \\ 1254\textbf{Output}. Zero the contents of $a$ \\ 1255\hline \\ 12561. $a.used \leftarrow 0$ \\ 12572. $a.sign \leftarrow$ MP\_ZPOS \\ 12583. for $n$ from 0 to $a.alloc - 1$ do \\ 1259\hspace{3mm}3.1 $a_n \leftarrow 0$ \\ 1260\hline 1261\end{tabular} 1262\end{center} 1263\caption{Algorithm mp\_zero} 1264\end{figure} 1265 1266\textbf{Algorithm mp\_zero.} 1267This algorithm simply resets a mp\_int to the default state. 1268 1269EXAM,bn_mp_zero.c 1270 1271After the function is completed, all of the digits are zeroed, the \textbf{used} count is zeroed and the 1272\textbf{sign} variable is set to \textbf{MP\_ZPOS}. 1273 1274\section{Sign Manipulation} 1275\subsection{Absolute Value} 1276With the mp\_int representation of an integer, calculating the absolute value is trivial. The mp\_abs algorithm will compute 1277the absolute value of an mp\_int. 1278 1279\begin{figure}[here] 1280\begin{center} 1281\begin{tabular}{l} 1282\hline Algorithm \textbf{mp\_abs}. \\ 1283\textbf{Input}. An mp\_int $a$ \\ 1284\textbf{Output}. Computes $b = \vert a \vert$ \\ 1285\hline \\ 12861. Copy $a$ to $b$. (\textit{mp\_copy}) \\ 12872. If the copy failed return(\textit{MP\_MEM}). \\ 12883. $b.sign \leftarrow MP\_ZPOS$ \\ 12894. Return(\textit{MP\_OKAY}) \\ 1290\hline 1291\end{tabular} 1292\end{center} 1293\caption{Algorithm mp\_abs} 1294\end{figure} 1295 1296\textbf{Algorithm mp\_abs.} 1297This algorithm computes the absolute of an mp\_int input. First it copies $a$ over $b$. This is an example of an 1298algorithm where the check in mp\_copy that determines if the source and destination are equal proves useful. This allows, 1299for instance, the developer to pass the same mp\_int as the source and destination to this function without addition 1300logic to handle it. 1301 1302EXAM,bn_mp_abs.c 1303 1304This fairly trivial algorithm first eliminates non--required duplications (line @27,a != b@) and then sets the 1305\textbf{sign} flag to \textbf{MP\_ZPOS}. 1306 1307\subsection{Integer Negation} 1308With the mp\_int representation of an integer, calculating the negation is also trivial. The mp\_neg algorithm will compute 1309the negative of an mp\_int input. 1310 1311\begin{figure}[here] 1312\begin{center} 1313\begin{tabular}{l} 1314\hline Algorithm \textbf{mp\_neg}. \\ 1315\textbf{Input}. An mp\_int $a$ \\ 1316\textbf{Output}. Computes $b = -a$ \\ 1317\hline \\ 13181. Copy $a$ to $b$. (\textit{mp\_copy}) \\ 13192. If the copy failed return(\textit{MP\_MEM}). \\ 13203. If $a.used = 0$ then return(\textit{MP\_OKAY}). \\ 13214. If $a.sign = MP\_ZPOS$ then do \\ 1322\hspace{3mm}4.1 $b.sign = MP\_NEG$. \\ 13235. else do \\ 1324\hspace{3mm}5.1 $b.sign = MP\_ZPOS$. \\ 13256. Return(\textit{MP\_OKAY}) \\ 1326\hline 1327\end{tabular} 1328\end{center} 1329\caption{Algorithm mp\_neg} 1330\end{figure} 1331 1332\textbf{Algorithm mp\_neg.} 1333This algorithm computes the negation of an input. First it copies $a$ over $b$. If $a$ has no used digits then 1334the algorithm returns immediately. Otherwise it flips the sign flag and stores the result in $b$. Note that if 1335$a$ had no digits then it must be positive by definition. Had step three been omitted then the algorithm would return 1336zero as negative. 1337 1338EXAM,bn_mp_neg.c 1339 1340Like mp\_abs() this function avoids non--required duplications (line @21,a != b@) and then sets the sign. We 1341have to make sure that only non--zero values get a \textbf{sign} of \textbf{MP\_NEG}. If the mp\_int is zero 1342than the \textbf{sign} is hard--coded to \textbf{MP\_ZPOS}. 1343 1344\section{Small Constants} 1345\subsection{Setting Small Constants} 1346Often a mp\_int must be set to a relatively small value such as $1$ or $2$. For these cases the mp\_set algorithm is useful. 1347 1348\newpage\begin{figure}[here] 1349\begin{center} 1350\begin{tabular}{l} 1351\hline Algorithm \textbf{mp\_set}. \\ 1352\textbf{Input}. An mp\_int $a$ and a digit $b$ \\ 1353\textbf{Output}. Make $a$ equivalent to $b$ \\ 1354\hline \\ 13551. Zero $a$ (\textit{mp\_zero}). \\ 13562. $a_0 \leftarrow b \mbox{ (mod }\beta\mbox{)}$ \\ 13573. $a.used \leftarrow \left \lbrace \begin{array}{ll} 1358 1 & \mbox{if }a_0 > 0 \\ 1359 0 & \mbox{if }a_0 = 0 1360 \end{array} \right .$ \\ 1361\hline 1362\end{tabular} 1363\end{center} 1364\caption{Algorithm mp\_set} 1365\end{figure} 1366 1367\textbf{Algorithm mp\_set.} 1368This algorithm sets a mp\_int to a small single digit value. Step number 1 ensures that the integer is reset to the default state. The 1369single digit is set (\textit{modulo $\beta$}) and the \textbf{used} count is adjusted accordingly. 1370 1371EXAM,bn_mp_set.c 1372 1373First we zero (line @21,mp_zero@) the mp\_int to make sure that the other members are initialized for a 1374small positive constant. mp\_zero() ensures that the \textbf{sign} is positive and the \textbf{used} count 1375is zero. Next we set the digit and reduce it modulo $\beta$ (line @22,MP_MASK@). After this step we have to 1376check if the resulting digit is zero or not. If it is not then we set the \textbf{used} count to one, otherwise 1377to zero. 1378 1379We can quickly reduce modulo $\beta$ since it is of the form $2^k$ and a quick binary AND operation with 1380$2^k - 1$ will perform the same operation. 1381 1382One important limitation of this function is that it will only set one digit. The size of a digit is not fixed, meaning source that uses 1383this function should take that into account. Only trivially small constants can be set using this function. 1384 1385\subsection{Setting Large Constants} 1386To overcome the limitations of the mp\_set algorithm the mp\_set\_int algorithm is ideal. It accepts a ``long'' 1387data type as input and will always treat it as a 32-bit integer. 1388 1389\begin{figure}[here] 1390\begin{center} 1391\begin{tabular}{l} 1392\hline Algorithm \textbf{mp\_set\_int}. \\ 1393\textbf{Input}. An mp\_int $a$ and a ``long'' integer $b$ \\ 1394\textbf{Output}. Make $a$ equivalent to $b$ \\ 1395\hline \\ 13961. Zero $a$ (\textit{mp\_zero}) \\ 13972. for $n$ from 0 to 7 do \\ 1398\hspace{3mm}2.1 $a \leftarrow a \cdot 16$ (\textit{mp\_mul2d}) \\ 1399\hspace{3mm}2.2 $u \leftarrow \lfloor b / 2^{4(7 - n)} \rfloor \mbox{ (mod }16\mbox{)}$\\ 1400\hspace{3mm}2.3 $a_0 \leftarrow a_0 + u$ \\ 1401\hspace{3mm}2.4 $a.used \leftarrow a.used + 1$ \\ 14023. Clamp excess used digits (\textit{mp\_clamp}) \\ 1403\hline 1404\end{tabular} 1405\end{center} 1406\caption{Algorithm mp\_set\_int} 1407\end{figure} 1408 1409\textbf{Algorithm mp\_set\_int.} 1410The algorithm performs eight iterations of a simple loop where in each iteration four bits from the source are added to the 1411mp\_int. Step 2.1 will multiply the current result by sixteen making room for four more bits in the less significant positions. In step 2.2 the 1412next four bits from the source are extracted and are added to the mp\_int. The \textbf{used} digit count is 1413incremented to reflect the addition. The \textbf{used} digit counter is incremented since if any of the leading digits were zero the mp\_int would have 1414zero digits used and the newly added four bits would be ignored. 1415 1416Excess zero digits are trimmed in steps 2.1 and 3 by using higher level algorithms mp\_mul2d and mp\_clamp. 1417 1418EXAM,bn_mp_set_int.c 1419 1420This function sets four bits of the number at a time to handle all practical \textbf{DIGIT\_BIT} sizes. The weird 1421addition on line @38,a->used@ ensures that the newly added in bits are added to the number of digits. While it may not 1422seem obvious as to why the digit counter does not grow exceedingly large it is because of the shift on line @27,mp_mul_2d@ 1423as well as the call to mp\_clamp() on line @40,mp_clamp@. Both functions will clamp excess leading digits which keeps 1424the number of used digits low. 1425 1426\section{Comparisons} 1427\subsection{Unsigned Comparisions} 1428Comparing a multiple precision integer is performed with the exact same algorithm used to compare two decimal numbers. For example, 1429to compare $1,234$ to $1,264$ the digits are extracted by their positions. That is we compare $1 \cdot 10^3 + 2 \cdot 10^2 + 3 \cdot 10^1 + 4 \cdot 10^0$ 1430to $1 \cdot 10^3 + 2 \cdot 10^2 + 6 \cdot 10^1 + 4 \cdot 10^0$ by comparing single digits at a time starting with the highest magnitude 1431positions. If any leading digit of one integer is greater than a digit in the same position of another integer then obviously it must be greater. 1432 1433The first comparision routine that will be developed is the unsigned magnitude compare which will perform a comparison based on the digits of two 1434mp\_int variables alone. It will ignore the sign of the two inputs. Such a function is useful when an absolute comparison is required or if the 1435signs are known to agree in advance. 1436 1437To facilitate working with the results of the comparison functions three constants are required. 1438 1439\begin{figure}[here] 1440\begin{center} 1441\begin{tabular}{|r|l|} 1442\hline \textbf{Constant} & \textbf{Meaning} \\ 1443\hline \textbf{MP\_GT} & Greater Than \\ 1444\hline \textbf{MP\_EQ} & Equal To \\ 1445\hline \textbf{MP\_LT} & Less Than \\ 1446\hline 1447\end{tabular} 1448\end{center} 1449\caption{Comparison Return Codes} 1450\end{figure} 1451 1452\begin{figure}[here] 1453\begin{center} 1454\begin{tabular}{l} 1455\hline Algorithm \textbf{mp\_cmp\_mag}. \\ 1456\textbf{Input}. Two mp\_ints $a$ and $b$. \\ 1457\textbf{Output}. Unsigned comparison results ($a$ to the left of $b$). \\ 1458\hline \\ 14591. If $a.used > b.used$ then return(\textit{MP\_GT}) \\ 14602. If $a.used < b.used$ then return(\textit{MP\_LT}) \\ 14613. for n from $a.used - 1$ to 0 do \\ 1462\hspace{+3mm}3.1 if $a_n > b_n$ then return(\textit{MP\_GT}) \\ 1463\hspace{+3mm}3.2 if $a_n < b_n$ then return(\textit{MP\_LT}) \\ 14644. Return(\textit{MP\_EQ}) \\ 1465\hline 1466\end{tabular} 1467\end{center} 1468\caption{Algorithm mp\_cmp\_mag} 1469\end{figure} 1470 1471\textbf{Algorithm mp\_cmp\_mag.} 1472By saying ``$a$ to the left of $b$'' it is meant that the comparison is with respect to $a$, that is if $a$ is greater than $b$ it will return 1473\textbf{MP\_GT} and similar with respect to when $a = b$ and $a < b$. The first two steps compare the number of digits used in both $a$ and $b$. 1474Obviously if the digit counts differ there would be an imaginary zero digit in the smaller number where the leading digit of the larger number is. 1475If both have the same number of digits than the actual digits themselves must be compared starting at the leading digit. 1476 1477By step three both inputs must have the same number of digits so its safe to start from either $a.used - 1$ or $b.used - 1$ and count down to 1478the zero'th digit. If after all of the digits have been compared, no difference is found, the algorithm returns \textbf{MP\_EQ}. 1479 1480EXAM,bn_mp_cmp_mag.c 1481 1482The two if statements (lines @24,if@ and @28,if@) compare the number of digits in the two inputs. These two are 1483performed before all of the digits are compared since it is a very cheap test to perform and can potentially save 1484considerable time. The implementation given is also not valid without those two statements. $b.alloc$ may be 1485smaller than $a.used$, meaning that undefined values will be read from $b$ past the end of the array of digits. 1486 1487 1488 1489\subsection{Signed Comparisons} 1490Comparing with sign considerations is also fairly critical in several routines (\textit{division for example}). Based on an unsigned magnitude 1491comparison a trivial signed comparison algorithm can be written. 1492 1493\begin{figure}[here] 1494\begin{center} 1495\begin{tabular}{l} 1496\hline Algorithm \textbf{mp\_cmp}. \\ 1497\textbf{Input}. Two mp\_ints $a$ and $b$ \\ 1498\textbf{Output}. Signed Comparison Results ($a$ to the left of $b$) \\ 1499\hline \\ 15001. if $a.sign = MP\_NEG$ and $b.sign = MP\_ZPOS$ then return(\textit{MP\_LT}) \\ 15012. if $a.sign = MP\_ZPOS$ and $b.sign = MP\_NEG$ then return(\textit{MP\_GT}) \\ 15023. if $a.sign = MP\_NEG$ then \\ 1503\hspace{+3mm}3.1 Return the unsigned comparison of $b$ and $a$ (\textit{mp\_cmp\_mag}) \\ 15044 Otherwise \\ 1505\hspace{+3mm}4.1 Return the unsigned comparison of $a$ and $b$ \\ 1506\hline 1507\end{tabular} 1508\end{center} 1509\caption{Algorithm mp\_cmp} 1510\end{figure} 1511 1512\textbf{Algorithm mp\_cmp.} 1513The first two steps compare the signs of the two inputs. If the signs do not agree then it can return right away with the appropriate 1514comparison code. When the signs are equal the digits of the inputs must be compared to determine the correct result. In step 1515three the unsigned comparision flips the order of the arguments since they are both negative. For instance, if $-a > -b$ then 1516$\vert a \vert < \vert b \vert$. Step number four will compare the two when they are both positive. 1517 1518EXAM,bn_mp_cmp.c 1519 1520The two if statements (lines @22,if@ and @26,if@) perform the initial sign comparison. If the signs are not the equal then which ever 1521has the positive sign is larger. The inputs are compared (line @30,if@) based on magnitudes. If the signs were both 1522negative then the unsigned comparison is performed in the opposite direction (line @31,mp_cmp_mag@). Otherwise, the signs are assumed to 1523be both positive and a forward direction unsigned comparison is performed. 1524 1525\section*{Exercises} 1526\begin{tabular}{cl} 1527$\left [ 2 \right ]$ & Modify algorithm mp\_set\_int to accept as input a variable length array of bits. \\ 1528 & \\ 1529$\left [ 3 \right ]$ & Give the probability that algorithm mp\_cmp\_mag will have to compare $k$ digits \\ 1530 & of two random digits (of equal magnitude) before a difference is found. \\ 1531 & \\ 1532$\left [ 1 \right ]$ & Suggest a simple method to speed up the implementation of mp\_cmp\_mag based \\ 1533 & on the observations made in the previous problem. \\ 1534 & 1535\end{tabular} 1536 1537\chapter{Basic Arithmetic} 1538\section{Introduction} 1539At this point algorithms for initialization, clearing, zeroing, copying, comparing and setting small constants have been 1540established. The next logical set of algorithms to develop are addition, subtraction and digit shifting algorithms. These 1541algorithms make use of the lower level algorithms and are the cruicial building block for the multiplication algorithms. It is very important 1542that these algorithms are highly optimized. On their own they are simple $O(n)$ algorithms but they can be called from higher level algorithms 1543which easily places them at $O(n^2)$ or even $O(n^3)$ work levels. 1544 1545MARK,SHIFTS 1546All of the algorithms within this chapter make use of the logical bit shift operations denoted by $<<$ and $>>$ for left and right 1547logical shifts respectively. A logical shift is analogous to sliding the decimal point of radix-10 representations. For example, the real 1548number $0.9345$ is equivalent to $93.45\%$ which is found by sliding the the decimal two places to the right (\textit{multiplying by $\beta^2 = 10^2$}). 1549Algebraically a binary logical shift is equivalent to a division or multiplication by a power of two. 1550For example, $a << k = a \cdot 2^k$ while $a >> k = \lfloor a/2^k \rfloor$. 1551 1552One significant difference between a logical shift and the way decimals are shifted is that digits below the zero'th position are removed 1553from the number. For example, consider $1101_2 >> 1$ using decimal notation this would produce $110.1_2$. However, with a logical shift the 1554result is $110_2$. 1555 1556\section{Addition and Subtraction} 1557In common twos complement fixed precision arithmetic negative numbers are easily represented by subtraction from the modulus. For example, with 32-bit integers 1558$a - b\mbox{ (mod }2^{32}\mbox{)}$ is the same as $a + (2^{32} - b) \mbox{ (mod }2^{32}\mbox{)}$ since $2^{32} \equiv 0 \mbox{ (mod }2^{32}\mbox{)}$. 1559As a result subtraction can be performed with a trivial series of logical operations and an addition. 1560 1561However, in multiple precision arithmetic negative numbers are not represented in the same way. Instead a sign flag is used to keep track of the 1562sign of the integer. As a result signed addition and subtraction are actually implemented as conditional usage of lower level addition or 1563subtraction algorithms with the sign fixed up appropriately. 1564 1565The lower level algorithms will add or subtract integers without regard to the sign flag. That is they will add or subtract the magnitude of 1566the integers respectively. 1567 1568\subsection{Low Level Addition} 1569An unsigned addition of multiple precision integers is performed with the same long-hand algorithm used to add decimal numbers. That is to add the 1570trailing digits first and propagate the resulting carry upwards. Since this is a lower level algorithm the name will have a ``s\_'' prefix. 1571Historically that convention stems from the MPI library where ``s\_'' stood for static functions that were hidden from the developer entirely. 1572 1573\newpage 1574\begin{figure}[!here] 1575\begin{center} 1576\begin{small} 1577\begin{tabular}{l} 1578\hline Algorithm \textbf{s\_mp\_add}. \\ 1579\textbf{Input}. Two mp\_ints $a$ and $b$ \\ 1580\textbf{Output}. The unsigned addition $c = \vert a \vert + \vert b \vert$. \\ 1581\hline \\ 15821. if $a.used > b.used$ then \\ 1583\hspace{+3mm}1.1 $min \leftarrow b.used$ \\ 1584\hspace{+3mm}1.2 $max \leftarrow a.used$ \\ 1585\hspace{+3mm}1.3 $x \leftarrow a$ \\ 15862. else \\ 1587\hspace{+3mm}2.1 $min \leftarrow a.used$ \\ 1588\hspace{+3mm}2.2 $max \leftarrow b.used$ \\ 1589\hspace{+3mm}2.3 $x \leftarrow b$ \\ 15903. If $c.alloc < max + 1$ then grow $c$ to hold at least $max + 1$ digits (\textit{mp\_grow}) \\ 15914. $oldused \leftarrow c.used$ \\ 15925. $c.used \leftarrow max + 1$ \\ 15936. $u \leftarrow 0$ \\ 15947. for $n$ from $0$ to $min - 1$ do \\ 1595\hspace{+3mm}7.1 $c_n \leftarrow a_n + b_n + u$ \\ 1596\hspace{+3mm}7.2 $u \leftarrow c_n >> lg(\beta)$ \\ 1597\hspace{+3mm}7.3 $c_n \leftarrow c_n \mbox{ (mod }\beta\mbox{)}$ \\ 15988. if $min \ne max$ then do \\ 1599\hspace{+3mm}8.1 for $n$ from $min$ to $max - 1$ do \\ 1600\hspace{+6mm}8.1.1 $c_n \leftarrow x_n + u$ \\ 1601\hspace{+6mm}8.1.2 $u \leftarrow c_n >> lg(\beta)$ \\ 1602\hspace{+6mm}8.1.3 $c_n \leftarrow c_n \mbox{ (mod }\beta\mbox{)}$ \\ 16039. $c_{max} \leftarrow u$ \\ 160410. if $olduse > max$ then \\ 1605\hspace{+3mm}10.1 for $n$ from $max + 1$ to $oldused - 1$ do \\ 1606\hspace{+6mm}10.1.1 $c_n \leftarrow 0$ \\ 160711. Clamp excess digits in $c$. (\textit{mp\_clamp}) \\ 160812. Return(\textit{MP\_OKAY}) \\ 1609\hline 1610\end{tabular} 1611\end{small} 1612\end{center} 1613\caption{Algorithm s\_mp\_add} 1614\end{figure} 1615 1616\textbf{Algorithm s\_mp\_add.} 1617This algorithm is loosely based on algorithm 14.7 of HAC \cite[pp. 594]{HAC} but has been extended to allow the inputs to have different magnitudes. 1618Coincidentally the description of algorithm A in Knuth \cite[pp. 266]{TAOCPV2} shares the same deficiency as the algorithm from \cite{HAC}. Even the 1619MIX pseudo machine code presented by Knuth \cite[pp. 266-267]{TAOCPV2} is incapable of handling inputs which are of different magnitudes. 1620 1621The first thing that has to be accomplished is to sort out which of the two inputs is the largest. The addition logic 1622will simply add all of the smallest input to the largest input and store that first part of the result in the 1623destination. Then it will apply a simpler addition loop to excess digits of the larger input. 1624 1625The first two steps will handle sorting the inputs such that $min$ and $max$ hold the digit counts of the two 1626inputs. The variable $x$ will be an mp\_int alias for the largest input or the second input $b$ if they have the 1627same number of digits. After the inputs are sorted the destination $c$ is grown as required to accomodate the sum 1628of the two inputs. The original \textbf{used} count of $c$ is copied and set to the new used count. 1629 1630At this point the first addition loop will go through as many digit positions that both inputs have. The carry 1631variable $\mu$ is set to zero outside the loop. Inside the loop an ``addition'' step requires three statements to produce 1632one digit of the summand. First 1633two digits from $a$ and $b$ are added together along with the carry $\mu$. The carry of this step is extracted and stored 1634in $\mu$ and finally the digit of the result $c_n$ is truncated within the range $0 \le c_n < \beta$. 1635 1636Now all of the digit positions that both inputs have in common have been exhausted. If $min \ne max$ then $x$ is an alias 1637for one of the inputs that has more digits. A simplified addition loop is then used to essentially copy the remaining digits 1638and the carry to the destination. 1639 1640The final carry is stored in $c_{max}$ and digits above $max$ upto $oldused$ are zeroed which completes the addition. 1641 1642 1643EXAM,bn_s_mp_add.c 1644 1645We first sort (lines @27,if@ to @35,}@) the inputs based on magnitude and determine the $min$ and $max$ variables. 1646Note that $x$ is a pointer to an mp\_int assigned to the largest input, in effect it is a local alias. Next we 1647grow the destination (@37,init@ to @42,}@) ensure that it can accomodate the result of the addition. 1648 1649Similar to the implementation of mp\_copy this function uses the braced code and local aliases coding style. The three aliases that are on 1650lines @56,tmpa@, @59,tmpb@ and @62,tmpc@ represent the two inputs and destination variables respectively. These aliases are used to ensure the 1651compiler does not have to dereference $a$, $b$ or $c$ (respectively) to access the digits of the respective mp\_int. 1652 1653The initial carry $u$ will be cleared (line @65,u = 0@), note that $u$ is of type mp\_digit which ensures type 1654compatibility within the implementation. The initial addition (line @66,for@ to @75,}@) adds digits from 1655both inputs until the smallest input runs out of digits. Similarly the conditional addition loop 1656(line @81,for@ to @90,}@) adds the remaining digits from the larger of the two inputs. The addition is finished 1657with the final carry being stored in $tmpc$ (line @94,tmpc++@). Note the ``++'' operator within the same expression. 1658After line @94,tmpc++@, $tmpc$ will point to the $c.used$'th digit of the mp\_int $c$. This is useful 1659for the next loop (line @97,for@ to @99,}@) which set any old upper digits to zero. 1660 1661\subsection{Low Level Subtraction} 1662The low level unsigned subtraction algorithm is very similar to the low level unsigned addition algorithm. The principle difference is that the 1663unsigned subtraction algorithm requires the result to be positive. That is when computing $a - b$ the condition $\vert a \vert \ge \vert b\vert$ must 1664be met for this algorithm to function properly. Keep in mind this low level algorithm is not meant to be used in higher level algorithms directly. 1665This algorithm as will be shown can be used to create functional signed addition and subtraction algorithms. 1666 1667MARK,GAMMA 1668 1669For this algorithm a new variable is required to make the description simpler. Recall from section 1.3.1 that a mp\_digit must be able to represent 1670the range $0 \le x < 2\beta$ for the algorithms to work correctly. However, it is allowable that a mp\_digit represent a larger range of values. For 1671this algorithm we will assume that the variable $\gamma$ represents the number of bits available in a 1672mp\_digit (\textit{this implies $2^{\gamma} > \beta$}). 1673 1674For example, the default for LibTomMath is to use a ``unsigned long'' for the mp\_digit ``type'' while $\beta = 2^{28}$. In ISO C an ``unsigned long'' 1675data type must be able to represent $0 \le x < 2^{32}$ meaning that in this case $\gamma \ge 32$. 1676 1677\newpage\begin{figure}[!here] 1678\begin{center} 1679\begin{small} 1680\begin{tabular}{l} 1681\hline Algorithm \textbf{s\_mp\_sub}. \\ 1682\textbf{Input}. Two mp\_ints $a$ and $b$ ($\vert a \vert \ge \vert b \vert$) \\ 1683\textbf{Output}. The unsigned subtraction $c = \vert a \vert - \vert b \vert$. \\ 1684\hline \\ 16851. $min \leftarrow b.used$ \\ 16862. $max \leftarrow a.used$ \\ 16873. If $c.alloc < max$ then grow $c$ to hold at least $max$ digits. (\textit{mp\_grow}) \\ 16884. $oldused \leftarrow c.used$ \\ 16895. $c.used \leftarrow max$ \\ 16906. $u \leftarrow 0$ \\ 16917. for $n$ from $0$ to $min - 1$ do \\ 1692\hspace{3mm}7.1 $c_n \leftarrow a_n - b_n - u$ \\ 1693\hspace{3mm}7.2 $u \leftarrow c_n >> (\gamma - 1)$ \\ 1694\hspace{3mm}7.3 $c_n \leftarrow c_n \mbox{ (mod }\beta\mbox{)}$ \\ 16958. if $min < max$ then do \\ 1696\hspace{3mm}8.1 for $n$ from $min$ to $max - 1$ do \\ 1697\hspace{6mm}8.1.1 $c_n \leftarrow a_n - u$ \\ 1698\hspace{6mm}8.1.2 $u \leftarrow c_n >> (\gamma - 1)$ \\ 1699\hspace{6mm}8.1.3 $c_n \leftarrow c_n \mbox{ (mod }\beta\mbox{)}$ \\ 17009. if $oldused > max$ then do \\ 1701\hspace{3mm}9.1 for $n$ from $max$ to $oldused - 1$ do \\ 1702\hspace{6mm}9.1.1 $c_n \leftarrow 0$ \\ 170310. Clamp excess digits of $c$. (\textit{mp\_clamp}). \\ 170411. Return(\textit{MP\_OKAY}). \\ 1705\hline 1706\end{tabular} 1707\end{small} 1708\end{center} 1709\caption{Algorithm s\_mp\_sub} 1710\end{figure} 1711 1712\textbf{Algorithm s\_mp\_sub.} 1713This algorithm performs the unsigned subtraction of two mp\_int variables under the restriction that the result must be positive. That is when 1714passing variables $a$ and $b$ the condition that $\vert a \vert \ge \vert b \vert$ must be met for the algorithm to function correctly. This 1715algorithm is loosely based on algorithm 14.9 \cite[pp. 595]{HAC} and is similar to algorithm S in \cite[pp. 267]{TAOCPV2} as well. As was the case 1716of the algorithm s\_mp\_add both other references lack discussion concerning various practical details such as when the inputs differ in magnitude. 1717 1718The initial sorting of the inputs is trivial in this algorithm since $a$ is guaranteed to have at least the same magnitude of $b$. Steps 1 and 2 1719set the $min$ and $max$ variables. Unlike the addition routine there is guaranteed to be no carry which means that the final result can be at 1720most $max$ digits in length as opposed to $max + 1$. Similar to the addition algorithm the \textbf{used} count of $c$ is copied locally and 1721set to the maximal count for the operation. 1722 1723The subtraction loop that begins on step seven is essentially the same as the addition loop of algorithm s\_mp\_add except single precision 1724subtraction is used instead. Note the use of the $\gamma$ variable to extract the carry (\textit{also known as the borrow}) within the subtraction 1725loops. Under the assumption that two's complement single precision arithmetic is used this will successfully extract the desired carry. 1726 1727For example, consider subtracting $0101_2$ from $0100_2$ where $\gamma = 4$ and $\beta = 2$. The least significant bit will force a carry upwards to 1728the third bit which will be set to zero after the borrow. After the very first bit has been subtracted $4 - 1 \equiv 0011_2$ will remain, When the 1729third bit of $0101_2$ is subtracted from the result it will cause another carry. In this case though the carry will be forced to propagate all the 1730way to the most significant bit. 1731 1732Recall that $\beta < 2^{\gamma}$. This means that if a carry does occur just before the $lg(\beta)$'th bit it will propagate all the way to the most 1733significant bit. Thus, the high order bits of the mp\_digit that are not part of the actual digit will either be all zero, or all one. All that 1734is needed is a single zero or one bit for the carry. Therefore a single logical shift right by $\gamma - 1$ positions is sufficient to extract the 1735carry. This method of carry extraction may seem awkward but the reason for it becomes apparent when the implementation is discussed. 1736 1737If $b$ has a smaller magnitude than $a$ then step 9 will force the carry and copy operation to propagate through the larger input $a$ into $c$. Step 173810 will ensure that any leading digits of $c$ above the $max$'th position are zeroed. 1739 1740EXAM,bn_s_mp_sub.c 1741 1742Like low level addition we ``sort'' the inputs. Except in this case the sorting is hardcoded 1743(lines @24,min@ and @25,max@). In reality the $min$ and $max$ variables are only aliases and are only 1744used to make the source code easier to read. Again the pointer alias optimization is used 1745within this algorithm. The aliases $tmpa$, $tmpb$ and $tmpc$ are initialized 1746(lines @42,tmpa@, @43,tmpb@ and @44,tmpc@) for $a$, $b$ and $c$ respectively. 1747 1748The first subtraction loop (lines @47,u = 0@ through @61,}@) subtract digits from both inputs until the smaller of 1749the two inputs has been exhausted. As remarked earlier there is an implementation reason for using the ``awkward'' 1750method of extracting the carry (line @57, >>@). The traditional method for extracting the carry would be to shift 1751by $lg(\beta)$ positions and logically AND the least significant bit. The AND operation is required because all of 1752the bits above the $\lg(\beta)$'th bit will be set to one after a carry occurs from subtraction. This carry 1753extraction requires two relatively cheap operations to extract the carry. The other method is to simply shift the 1754most significant bit to the least significant bit thus extracting the carry with a single cheap operation. This 1755optimization only works on twos compliment machines which is a safe assumption to make. 1756 1757If $a$ has a larger magnitude than $b$ an additional loop (lines @64,for@ through @73,}@) is required to propagate 1758the carry through $a$ and copy the result to $c$. 1759 1760\subsection{High Level Addition} 1761Now that both lower level addition and subtraction algorithms have been established an effective high level signed addition algorithm can be 1762established. This high level addition algorithm will be what other algorithms and developers will use to perform addition of mp\_int data 1763types. 1764 1765Recall from section 5.2 that an mp\_int represents an integer with an unsigned mantissa (\textit{the array of digits}) and a \textbf{sign} 1766flag. A high level addition is actually performed as a series of eight separate cases which can be optimized down to three unique cases. 1767 1768\begin{figure}[!here] 1769\begin{center} 1770\begin{tabular}{l} 1771\hline Algorithm \textbf{mp\_add}. \\ 1772\textbf{Input}. Two mp\_ints $a$ and $b$ \\ 1773\textbf{Output}. The signed addition $c = a + b$. \\ 1774\hline \\ 17751. if $a.sign = b.sign$ then do \\ 1776\hspace{3mm}1.1 $c.sign \leftarrow a.sign$ \\ 1777\hspace{3mm}1.2 $c \leftarrow \vert a \vert + \vert b \vert$ (\textit{s\_mp\_add})\\ 17782. else do \\ 1779\hspace{3mm}2.1 if $\vert a \vert < \vert b \vert$ then do (\textit{mp\_cmp\_mag}) \\ 1780\hspace{6mm}2.1.1 $c.sign \leftarrow b.sign$ \\ 1781\hspace{6mm}2.1.2 $c \leftarrow \vert b \vert - \vert a \vert$ (\textit{s\_mp\_sub}) \\ 1782\hspace{3mm}2.2 else do \\ 1783\hspace{6mm}2.2.1 $c.sign \leftarrow a.sign$ \\ 1784\hspace{6mm}2.2.2 $c \leftarrow \vert a \vert - \vert b \vert$ \\ 17853. Return(\textit{MP\_OKAY}). \\ 1786\hline 1787\end{tabular} 1788\end{center} 1789\caption{Algorithm mp\_add} 1790\end{figure} 1791 1792\textbf{Algorithm mp\_add.} 1793This algorithm performs the signed addition of two mp\_int variables. There is no reference algorithm to draw upon from 1794either \cite{TAOCPV2} or \cite{HAC} since they both only provide unsigned operations. The algorithm is fairly 1795straightforward but restricted since subtraction can only produce positive results. 1796 1797\begin{figure}[here] 1798\begin{small} 1799\begin{center} 1800\begin{tabular}{|c|c|c|c|c|} 1801\hline \textbf{Sign of $a$} & \textbf{Sign of $b$} & \textbf{$\vert a \vert > \vert b \vert $} & \textbf{Unsigned Operation} & \textbf{Result Sign Flag} \\ 1802\hline $+$ & $+$ & Yes & $c = a + b$ & $a.sign$ \\ 1803\hline $+$ & $+$ & No & $c = a + b$ & $a.sign$ \\ 1804\hline $-$ & $-$ & Yes & $c = a + b$ & $a.sign$ \\ 1805\hline $-$ & $-$ & No & $c = a + b$ & $a.sign$ \\ 1806\hline &&&&\\ 1807 1808\hline $+$ & $-$ & No & $c = b - a$ & $b.sign$ \\ 1809\hline $-$ & $+$ & No & $c = b - a$ & $b.sign$ \\ 1810 1811\hline &&&&\\ 1812 1813\hline $+$ & $-$ & Yes & $c = a - b$ & $a.sign$ \\ 1814\hline $-$ & $+$ & Yes & $c = a - b$ & $a.sign$ \\ 1815 1816\hline 1817\end{tabular} 1818\end{center} 1819\end{small} 1820\caption{Addition Guide Chart} 1821\label{fig:AddChart} 1822\end{figure} 1823 1824Figure~\ref{fig:AddChart} lists all of the eight possible input combinations and is sorted to show that only three 1825specific cases need to be handled. The return code of the unsigned operations at step 1.2, 2.1.2 and 2.2.2 are 1826forwarded to step three to check for errors. This simplifies the description of the algorithm considerably and best 1827follows how the implementation actually was achieved. 1828 1829Also note how the \textbf{sign} is set before the unsigned addition or subtraction is performed. Recall from the descriptions of algorithms 1830s\_mp\_add and s\_mp\_sub that the mp\_clamp function is used at the end to trim excess digits. The mp\_clamp algorithm will set the \textbf{sign} 1831to \textbf{MP\_ZPOS} when the \textbf{used} digit count reaches zero. 1832 1833For example, consider performing $-a + a$ with algorithm mp\_add. By the description of the algorithm the sign is set to \textbf{MP\_NEG} which would 1834produce a result of $-0$. However, since the sign is set first then the unsigned addition is performed the subsequent usage of algorithm mp\_clamp 1835within algorithm s\_mp\_add will force $-0$ to become $0$. 1836 1837EXAM,bn_mp_add.c 1838 1839The source code follows the algorithm fairly closely. The most notable new source code addition is the usage of the $res$ integer variable which 1840is used to pass result of the unsigned operations forward. Unlike in the algorithm, the variable $res$ is merely returned as is without 1841explicitly checking it and returning the constant \textbf{MP\_OKAY}. The observation is this algorithm will succeed or fail only if the lower 1842level functions do so. Returning their return code is sufficient. 1843 1844\subsection{High Level Subtraction} 1845The high level signed subtraction algorithm is essentially the same as the high level signed addition algorithm. 1846 1847\newpage\begin{figure}[!here] 1848\begin{center} 1849\begin{tabular}{l} 1850\hline Algorithm \textbf{mp\_sub}. \\ 1851\textbf{Input}. Two mp\_ints $a$ and $b$ \\ 1852\textbf{Output}. The signed subtraction $c = a - b$. \\ 1853\hline \\ 18541. if $a.sign \ne b.sign$ then do \\ 1855\hspace{3mm}1.1 $c.sign \leftarrow a.sign$ \\ 1856\hspace{3mm}1.2 $c \leftarrow \vert a \vert + \vert b \vert$ (\textit{s\_mp\_add}) \\ 18572. else do \\ 1858\hspace{3mm}2.1 if $\vert a \vert \ge \vert b \vert$ then do (\textit{mp\_cmp\_mag}) \\ 1859\hspace{6mm}2.1.1 $c.sign \leftarrow a.sign$ \\ 1860\hspace{6mm}2.1.2 $c \leftarrow \vert a \vert - \vert b \vert$ (\textit{s\_mp\_sub}) \\ 1861\hspace{3mm}2.2 else do \\ 1862\hspace{6mm}2.2.1 $c.sign \leftarrow \left \lbrace \begin{array}{ll} 1863 MP\_ZPOS & \mbox{if }a.sign = MP\_NEG \\ 1864 MP\_NEG & \mbox{otherwise} \\ 1865 \end{array} \right .$ \\ 1866\hspace{6mm}2.2.2 $c \leftarrow \vert b \vert - \vert a \vert$ \\ 18673. Return(\textit{MP\_OKAY}). \\ 1868\hline 1869\end{tabular} 1870\end{center} 1871\caption{Algorithm mp\_sub} 1872\end{figure} 1873 1874\textbf{Algorithm mp\_sub.} 1875This algorithm performs the signed subtraction of two inputs. Similar to algorithm mp\_add there is no reference in either \cite{TAOCPV2} or 1876\cite{HAC}. Also this algorithm is restricted by algorithm s\_mp\_sub. Chart \ref{fig:SubChart} lists the eight possible inputs and 1877the operations required. 1878 1879\begin{figure}[!here] 1880\begin{small} 1881\begin{center} 1882\begin{tabular}{|c|c|c|c|c|} 1883\hline \textbf{Sign of $a$} & \textbf{Sign of $b$} & \textbf{$\vert a \vert \ge \vert b \vert $} & \textbf{Unsigned Operation} & \textbf{Result Sign Flag} \\ 1884\hline $+$ & $-$ & Yes & $c = a + b$ & $a.sign$ \\ 1885\hline $+$ & $-$ & No & $c = a + b$ & $a.sign$ \\ 1886\hline $-$ & $+$ & Yes & $c = a + b$ & $a.sign$ \\ 1887\hline $-$ & $+$ & No & $c = a + b$ & $a.sign$ \\ 1888\hline &&&& \\ 1889\hline $+$ & $+$ & Yes & $c = a - b$ & $a.sign$ \\ 1890\hline $-$ & $-$ & Yes & $c = a - b$ & $a.sign$ \\ 1891\hline &&&& \\ 1892\hline $+$ & $+$ & No & $c = b - a$ & $\mbox{opposite of }a.sign$ \\ 1893\hline $-$ & $-$ & No & $c = b - a$ & $\mbox{opposite of }a.sign$ \\ 1894\hline 1895\end{tabular} 1896\end{center} 1897\end{small} 1898\caption{Subtraction Guide Chart} 1899\label{fig:SubChart} 1900\end{figure} 1901 1902Similar to the case of algorithm mp\_add the \textbf{sign} is set first before the unsigned addition or subtraction. That is to prevent the 1903algorithm from producing $-a - -a = -0$ as a result. 1904 1905EXAM,bn_mp_sub.c 1906 1907Much like the implementation of algorithm mp\_add the variable $res$ is used to catch the return code of the unsigned addition or subtraction operations 1908and forward it to the end of the function. On line @38, != MP_LT@ the ``not equal to'' \textbf{MP\_LT} expression is used to emulate a 1909``greater than or equal to'' comparison. 1910 1911\section{Bit and Digit Shifting} 1912MARK,POLY 1913It is quite common to think of a multiple precision integer as a polynomial in $x$, that is $y = f(\beta)$ where $f(x) = \sum_{i=0}^{n-1} a_i x^i$. 1914This notation arises within discussion of Montgomery and Diminished Radix Reduction as well as Karatsuba multiplication and squaring. 1915 1916In order to facilitate operations on polynomials in $x$ as above a series of simple ``digit'' algorithms have to be established. That is to shift 1917the digits left or right as well to shift individual bits of the digits left and right. It is important to note that not all ``shift'' operations 1918are on radix-$\beta$ digits. 1919 1920\subsection{Multiplication by Two} 1921 1922In a binary system where the radix is a power of two multiplication by two not only arises often in other algorithms it is a fairly efficient 1923operation to perform. A single precision logical shift left is sufficient to multiply a single digit by two. 1924 1925\newpage\begin{figure}[!here] 1926\begin{small} 1927\begin{center} 1928\begin{tabular}{l} 1929\hline Algorithm \textbf{mp\_mul\_2}. \\ 1930\textbf{Input}. One mp\_int $a$ \\ 1931\textbf{Output}. $b = 2a$. \\ 1932\hline \\ 19331. If $b.alloc < a.used + 1$ then grow $b$ to hold $a.used + 1$ digits. (\textit{mp\_grow}) \\ 19342. $oldused \leftarrow b.used$ \\ 19353. $b.used \leftarrow a.used$ \\ 19364. $r \leftarrow 0$ \\ 19375. for $n$ from 0 to $a.used - 1$ do \\ 1938\hspace{3mm}5.1 $rr \leftarrow a_n >> (lg(\beta) - 1)$ \\ 1939\hspace{3mm}5.2 $b_n \leftarrow (a_n << 1) + r \mbox{ (mod }\beta\mbox{)}$ \\ 1940\hspace{3mm}5.3 $r \leftarrow rr$ \\ 19416. If $r \ne 0$ then do \\ 1942\hspace{3mm}6.1 $b_{n + 1} \leftarrow r$ \\ 1943\hspace{3mm}6.2 $b.used \leftarrow b.used + 1$ \\ 19447. If $b.used < oldused - 1$ then do \\ 1945\hspace{3mm}7.1 for $n$ from $b.used$ to $oldused - 1$ do \\ 1946\hspace{6mm}7.1.1 $b_n \leftarrow 0$ \\ 19478. $b.sign \leftarrow a.sign$ \\ 19489. Return(\textit{MP\_OKAY}).\\ 1949\hline 1950\end{tabular} 1951\end{center} 1952\end{small} 1953\caption{Algorithm mp\_mul\_2} 1954\end{figure} 1955 1956\textbf{Algorithm mp\_mul\_2.} 1957This algorithm will quickly multiply a mp\_int by two provided $\beta$ is a power of two. Neither \cite{TAOCPV2} nor \cite{HAC} describe such 1958an algorithm despite the fact it arises often in other algorithms. The algorithm is setup much like the lower level algorithm s\_mp\_add since 1959it is for all intents and purposes equivalent to the operation $b = \vert a \vert + \vert a \vert$. 1960 1961Step 1 and 2 grow the input as required to accomodate the maximum number of \textbf{used} digits in the result. The initial \textbf{used} count 1962is set to $a.used$ at step 4. Only if there is a final carry will the \textbf{used} count require adjustment. 1963 1964Step 6 is an optimization implementation of the addition loop for this specific case. That is since the two values being added together 1965are the same there is no need to perform two reads from the digits of $a$. Step 6.1 performs a single precision shift on the current digit $a_n$ to 1966obtain what will be the carry for the next iteration. Step 6.2 calculates the $n$'th digit of the result as single precision shift of $a_n$ plus 1967the previous carry. Recall from ~SHIFTS~ that $a_n << 1$ is equivalent to $a_n \cdot 2$. An iteration of the addition loop is finished with 1968forwarding the carry to the next iteration. 1969 1970Step 7 takes care of any final carry by setting the $a.used$'th digit of the result to the carry and augmenting the \textbf{used} count of $b$. 1971Step 8 clears any leading digits of $b$ in case it originally had a larger magnitude than $a$. 1972 1973EXAM,bn_mp_mul_2.c 1974 1975This implementation is essentially an optimized implementation of s\_mp\_add for the case of doubling an input. The only noteworthy difference 1976is the use of the logical shift operator on line @52,<<@ to perform a single precision doubling. 1977 1978\subsection{Division by Two} 1979A division by two can just as easily be accomplished with a logical shift right as multiplication by two can be with a logical shift left. 1980 1981\newpage\begin{figure}[!here] 1982\begin{small} 1983\begin{center} 1984\begin{tabular}{l} 1985\hline Algorithm \textbf{mp\_div\_2}. \\ 1986\textbf{Input}. One mp\_int $a$ \\ 1987\textbf{Output}. $b = a/2$. \\ 1988\hline \\ 19891. If $b.alloc < a.used$ then grow $b$ to hold $a.used$ digits. (\textit{mp\_grow}) \\ 19902. If the reallocation failed return(\textit{MP\_MEM}). \\ 19913. $oldused \leftarrow b.used$ \\ 19924. $b.used \leftarrow a.used$ \\ 19935. $r \leftarrow 0$ \\ 19946. for $n$ from $b.used - 1$ to $0$ do \\ 1995\hspace{3mm}6.1 $rr \leftarrow a_n \mbox{ (mod }2\mbox{)}$\\ 1996\hspace{3mm}6.2 $b_n \leftarrow (a_n >> 1) + (r << (lg(\beta) - 1)) \mbox{ (mod }\beta\mbox{)}$ \\ 1997\hspace{3mm}6.3 $r \leftarrow rr$ \\ 19987. If $b.used < oldused - 1$ then do \\ 1999\hspace{3mm}7.1 for $n$ from $b.used$ to $oldused - 1$ do \\ 2000\hspace{6mm}7.1.1 $b_n \leftarrow 0$ \\ 20018. $b.sign \leftarrow a.sign$ \\ 20029. Clamp excess digits of $b$. (\textit{mp\_clamp}) \\ 200310. Return(\textit{MP\_OKAY}).\\ 2004\hline 2005\end{tabular} 2006\end{center} 2007\end{small} 2008\caption{Algorithm mp\_div\_2} 2009\end{figure} 2010 2011\textbf{Algorithm mp\_div\_2.} 2012This algorithm will divide an mp\_int by two using logical shifts to the right. Like mp\_mul\_2 it uses a modified low level addition 2013core as the basis of the algorithm. Unlike mp\_mul\_2 the shift operations work from the leading digit to the trailing digit. The algorithm 2014could be written to work from the trailing digit to the leading digit however, it would have to stop one short of $a.used - 1$ digits to prevent 2015reading past the end of the array of digits. 2016 2017Essentially the loop at step 6 is similar to that of mp\_mul\_2 except the logical shifts go in the opposite direction and the carry is at the 2018least significant bit not the most significant bit. 2019 2020EXAM,bn_mp_div_2.c 2021 2022\section{Polynomial Basis Operations} 2023Recall from ~POLY~ that any integer can be represented as a polynomial in $x$ as $y = f(\beta)$. Such a representation is also known as 2024the polynomial basis \cite[pp. 48]{ROSE}. Given such a notation a multiplication or division by $x$ amounts to shifting whole digits a single 2025place. The need for such operations arises in several other higher level algorithms such as Barrett and Montgomery reduction, integer 2026division and Karatsuba multiplication. 2027 2028Converting from an array of digits to polynomial basis is very simple. Consider the integer $y \equiv (a_2, a_1, a_0)_{\beta}$ and recall that 2029$y = \sum_{i=0}^{2} a_i \beta^i$. Simply replace $\beta$ with $x$ and the expression is in polynomial basis. For example, $f(x) = 8x + 9$ is the 2030polynomial basis representation for $89$ using radix ten. That is, $f(10) = 8(10) + 9 = 89$. 2031 2032\subsection{Multiplication by $x$} 2033 2034Given a polynomial in $x$ such as $f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_0$ multiplying by $x$ amounts to shifting the coefficients up one 2035degree. In this case $f(x) \cdot x = a_n x^{n+1} + a_{n-1} x^n + ... + a_0 x$. From a scalar basis point of view multiplying by $x$ is equivalent to 2036multiplying by the integer $\beta$. 2037 2038\newpage\begin{figure}[!here] 2039\begin{small} 2040\begin{center} 2041\begin{tabular}{l} 2042\hline Algorithm \textbf{mp\_lshd}. \\ 2043\textbf{Input}. One mp\_int $a$ and an integer $b$ \\ 2044\textbf{Output}. $a \leftarrow a \cdot \beta^b$ (equivalent to multiplication by $x^b$). \\ 2045\hline \\ 20461. If $b \le 0$ then return(\textit{MP\_OKAY}). \\ 20472. If $a.alloc < a.used + b$ then grow $a$ to at least $a.used + b$ digits. (\textit{mp\_grow}). \\ 20483. If the reallocation failed return(\textit{MP\_MEM}). \\ 20494. $a.used \leftarrow a.used + b$ \\ 20505. $i \leftarrow a.used - 1$ \\ 20516. $j \leftarrow a.used - 1 - b$ \\ 20527. for $n$ from $a.used - 1$ to $b$ do \\ 2053\hspace{3mm}7.1 $a_{i} \leftarrow a_{j}$ \\ 2054\hspace{3mm}7.2 $i \leftarrow i - 1$ \\ 2055\hspace{3mm}7.3 $j \leftarrow j - 1$ \\ 20568. for $n$ from 0 to $b - 1$ do \\ 2057\hspace{3mm}8.1 $a_n \leftarrow 0$ \\ 20589. Return(\textit{MP\_OKAY}). \\ 2059\hline 2060\end{tabular} 2061\end{center} 2062\end{small} 2063\caption{Algorithm mp\_lshd} 2064\end{figure} 2065 2066\textbf{Algorithm mp\_lshd.} 2067This algorithm multiplies an mp\_int by the $b$'th power of $x$. This is equivalent to multiplying by $\beta^b$. The algorithm differs 2068from the other algorithms presented so far as it performs the operation in place instead storing the result in a separate location. The 2069motivation behind this change is due to the way this function is typically used. Algorithms such as mp\_add store the result in an optionally 2070different third mp\_int because the original inputs are often still required. Algorithm mp\_lshd (\textit{and similarly algorithm mp\_rshd}) is 2071typically used on values where the original value is no longer required. The algorithm will return success immediately if 2072$b \le 0$ since the rest of algorithm is only valid when $b > 0$. 2073 2074First the destination $a$ is grown as required to accomodate the result. The counters $i$ and $j$ are used to form a \textit{sliding window} over 2075the digits of $a$ of length $b$. The head of the sliding window is at $i$ (\textit{the leading digit}) and the tail at $j$ (\textit{the trailing digit}). 2076The loop on step 7 copies the digit from the tail to the head. In each iteration the window is moved down one digit. The last loop on 2077step 8 sets the lower $b$ digits to zero. 2078 2079\newpage 2080FIGU,sliding_window,Sliding Window Movement 2081 2082EXAM,bn_mp_lshd.c 2083 2084The if statement (line @24,if@) ensures that the $b$ variable is greater than zero since we do not interpret negative 2085shift counts properly. The \textbf{used} count is incremented by $b$ before the copy loop begins. This elminates 2086the need for an additional variable in the for loop. The variable $top$ (line @42,top@) is an alias 2087for the leading digit while $bottom$ (line @45,bottom@) is an alias for the trailing edge. The aliases form a 2088window of exactly $b$ digits over the input. 2089 2090\subsection{Division by $x$} 2091 2092Division by powers of $x$ is easily achieved by shifting the digits right and removing any that will end up to the right of the zero'th digit. 2093 2094\newpage\begin{figure}[!here] 2095\begin{small} 2096\begin{center} 2097\begin{tabular}{l} 2098\hline Algorithm \textbf{mp\_rshd}. \\ 2099\textbf{Input}. One mp\_int $a$ and an integer $b$ \\ 2100\textbf{Output}. $a \leftarrow a / \beta^b$ (Divide by $x^b$). \\ 2101\hline \\ 21021. If $b \le 0$ then return. \\ 21032. If $a.used \le b$ then do \\ 2104\hspace{3mm}2.1 Zero $a$. (\textit{mp\_zero}). \\ 2105\hspace{3mm}2.2 Return. \\ 21063. $i \leftarrow 0$ \\ 21074. $j \leftarrow b$ \\ 21085. for $n$ from 0 to $a.used - b - 1$ do \\ 2109\hspace{3mm}5.1 $a_i \leftarrow a_j$ \\ 2110\hspace{3mm}5.2 $i \leftarrow i + 1$ \\ 2111\hspace{3mm}5.3 $j \leftarrow j + 1$ \\ 21126. for $n$ from $a.used - b$ to $a.used - 1$ do \\ 2113\hspace{3mm}6.1 $a_n \leftarrow 0$ \\ 21147. $a.used \leftarrow a.used - b$ \\ 21158. Return. \\ 2116\hline 2117\end{tabular} 2118\end{center} 2119\end{small} 2120\caption{Algorithm mp\_rshd} 2121\end{figure} 2122 2123\textbf{Algorithm mp\_rshd.} 2124This algorithm divides the input in place by the $b$'th power of $x$. It is analogous to dividing by a $\beta^b$ but much quicker since 2125it does not require single precision division. This algorithm does not actually return an error code as it cannot fail. 2126 2127If the input $b$ is less than one the algorithm quickly returns without performing any work. If the \textbf{used} count is less than or equal 2128to the shift count $b$ then it will simply zero the input and return. 2129 2130After the trivial cases of inputs have been handled the sliding window is setup. Much like the case of algorithm mp\_lshd a sliding window that 2131is $b$ digits wide is used to copy the digits. Unlike mp\_lshd the window slides in the opposite direction from the trailing to the leading digit. 2132Also the digits are copied from the leading to the trailing edge. 2133 2134Once the window copy is complete the upper digits must be zeroed and the \textbf{used} count decremented. 2135 2136EXAM,bn_mp_rshd.c 2137 2138The only noteworthy element of this routine is the lack of a return type since it cannot fail. Like mp\_lshd() we 2139form a sliding window except we copy in the other direction. After the window (line @59,for (;@) we then zero 2140the upper digits of the input to make sure the result is correct. 2141 2142\section{Powers of Two} 2143 2144Now that algorithms for moving single bits as well as whole digits exist algorithms for moving the ``in between'' distances are required. For 2145example, to quickly multiply by $2^k$ for any $k$ without using a full multiplier algorithm would prove useful. Instead of performing single 2146shifts $k$ times to achieve a multiplication by $2^{\pm k}$ a mixture of whole digit shifting and partial digit shifting is employed. 2147 2148\subsection{Multiplication by Power of Two} 2149 2150\newpage\begin{figure}[!here] 2151\begin{small} 2152\begin{center} 2153\begin{tabular}{l} 2154\hline Algorithm \textbf{mp\_mul\_2d}. \\ 2155\textbf{Input}. One mp\_int $a$ and an integer $b$ \\ 2156\textbf{Output}. $c \leftarrow a \cdot 2^b$. \\ 2157\hline \\ 21581. $c \leftarrow a$. (\textit{mp\_copy}) \\ 21592. If $c.alloc < c.used + \lfloor b / lg(\beta) \rfloor + 2$ then grow $c$ accordingly. \\ 21603. If the reallocation failed return(\textit{MP\_MEM}). \\ 21614. If $b \ge lg(\beta)$ then \\ 2162\hspace{3mm}4.1 $c \leftarrow c \cdot \beta^{\lfloor b / lg(\beta) \rfloor}$ (\textit{mp\_lshd}). \\ 2163\hspace{3mm}4.2 If step 4.1 failed return(\textit{MP\_MEM}). \\ 21645. $d \leftarrow b \mbox{ (mod }lg(\beta)\mbox{)}$ \\ 21656. If $d \ne 0$ then do \\ 2166\hspace{3mm}6.1 $mask \leftarrow 2^d$ \\ 2167\hspace{3mm}6.2 $r \leftarrow 0$ \\ 2168\hspace{3mm}6.3 for $n$ from $0$ to $c.used - 1$ do \\ 2169\hspace{6mm}6.3.1 $rr \leftarrow c_n >> (lg(\beta) - d) \mbox{ (mod }mask\mbox{)}$ \\ 2170\hspace{6mm}6.3.2 $c_n \leftarrow (c_n << d) + r \mbox{ (mod }\beta\mbox{)}$ \\ 2171\hspace{6mm}6.3.3 $r \leftarrow rr$ \\ 2172\hspace{3mm}6.4 If $r > 0$ then do \\ 2173\hspace{6mm}6.4.1 $c_{c.used} \leftarrow r$ \\ 2174\hspace{6mm}6.4.2 $c.used \leftarrow c.used + 1$ \\ 21757. Return(\textit{MP\_OKAY}). \\ 2176\hline 2177\end{tabular} 2178\end{center} 2179\end{small} 2180\caption{Algorithm mp\_mul\_2d} 2181\end{figure} 2182 2183\textbf{Algorithm mp\_mul\_2d.} 2184This algorithm multiplies $a$ by $2^b$ and stores the result in $c$. The algorithm uses algorithm mp\_lshd and a derivative of algorithm mp\_mul\_2 to 2185quickly compute the product. 2186 2187First the algorithm will multiply $a$ by $x^{\lfloor b / lg(\beta) \rfloor}$ which will ensure that the remainder multiplicand is less than 2188$\beta$. For example, if $b = 37$ and $\beta = 2^{28}$ then this step will multiply by $x$ leaving a multiplication by $2^{37 - 28} = 2^{9}$ 2189left. 2190 2191After the digits have been shifted appropriately at most $lg(\beta) - 1$ shifts are left to perform. Step 5 calculates the number of remaining shifts 2192required. If it is non-zero a modified shift loop is used to calculate the remaining product. 2193Essentially the loop is a generic version of algorithm mp\_mul\_2 designed to handle any shift count in the range $1 \le x < lg(\beta)$. The $mask$ 2194variable is used to extract the upper $d$ bits to form the carry for the next iteration. 2195 2196This algorithm is loosely measured as a $O(2n)$ algorithm which means that if the input is $n$-digits that it takes $2n$ ``time'' to 2197complete. It is possible to optimize this algorithm down to a $O(n)$ algorithm at a cost of making the algorithm slightly harder to follow. 2198 2199EXAM,bn_mp_mul_2d.c 2200 2201The shifting is performed in--place which means the first step (line @24,a != c@) is to copy the input to the 2202destination. We avoid calling mp\_copy() by making sure the mp\_ints are different. The destination then 2203has to be grown (line @31,grow@) to accomodate the result. 2204 2205If the shift count $b$ is larger than $lg(\beta)$ then a call to mp\_lshd() is used to handle all of the multiples 2206of $lg(\beta)$. Leaving only a remaining shift of $lg(\beta) - 1$ or fewer bits left. Inside the actual shift 2207loop (lines @45,if@ to @76,}@) we make use of pre--computed values $shift$ and $mask$. These are used to 2208extract the carry bit(s) to pass into the next iteration of the loop. The $r$ and $rr$ variables form a 2209chain between consecutive iterations to propagate the carry. 2210 2211\subsection{Division by Power of Two} 2212 2213\newpage\begin{figure}[!here] 2214\begin{small} 2215\begin{center} 2216\begin{tabular}{l} 2217\hline Algorithm \textbf{mp\_div\_2d}. \\ 2218\textbf{Input}. One mp\_int $a$ and an integer $b$ \\ 2219\textbf{Output}. $c \leftarrow \lfloor a / 2^b \rfloor, d \leftarrow a \mbox{ (mod }2^b\mbox{)}$. \\ 2220\hline \\ 22211. If $b \le 0$ then do \\ 2222\hspace{3mm}1.1 $c \leftarrow a$ (\textit{mp\_copy}) \\ 2223\hspace{3mm}1.2 $d \leftarrow 0$ (\textit{mp\_zero}) \\ 2224\hspace{3mm}1.3 Return(\textit{MP\_OKAY}). \\ 22252. $c \leftarrow a$ \\ 22263. $d \leftarrow a \mbox{ (mod }2^b\mbox{)}$ (\textit{mp\_mod\_2d}) \\ 22274. If $b \ge lg(\beta)$ then do \\ 2228\hspace{3mm}4.1 $c \leftarrow \lfloor c/\beta^{\lfloor b/lg(\beta) \rfloor} \rfloor$ (\textit{mp\_rshd}). \\ 22295. $k \leftarrow b \mbox{ (mod }lg(\beta)\mbox{)}$ \\ 22306. If $k \ne 0$ then do \\ 2231\hspace{3mm}6.1 $mask \leftarrow 2^k$ \\ 2232\hspace{3mm}6.2 $r \leftarrow 0$ \\ 2233\hspace{3mm}6.3 for $n$ from $c.used - 1$ to $0$ do \\ 2234\hspace{6mm}6.3.1 $rr \leftarrow c_n \mbox{ (mod }mask\mbox{)}$ \\ 2235\hspace{6mm}6.3.2 $c_n \leftarrow (c_n >> k) + (r << (lg(\beta) - k))$ \\ 2236\hspace{6mm}6.3.3 $r \leftarrow rr$ \\ 22377. Clamp excess digits of $c$. (\textit{mp\_clamp}) \\ 22388. Return(\textit{MP\_OKAY}). \\ 2239\hline 2240\end{tabular} 2241\end{center} 2242\end{small} 2243\caption{Algorithm mp\_div\_2d} 2244\end{figure} 2245 2246\textbf{Algorithm mp\_div\_2d.} 2247This algorithm will divide an input $a$ by $2^b$ and produce the quotient and remainder. The algorithm is designed much like algorithm 2248mp\_mul\_2d by first using whole digit shifts then single precision shifts. This algorithm will also produce the remainder of the division 2249by using algorithm mp\_mod\_2d. 2250 2251EXAM,bn_mp_div_2d.c 2252 2253The implementation of algorithm mp\_div\_2d is slightly different than the algorithm specifies. The remainder $d$ may be optionally 2254ignored by passing \textbf{NULL} as the pointer to the mp\_int variable. The temporary mp\_int variable $t$ is used to hold the 2255result of the remainder operation until the end. This allows $d$ and $a$ to represent the same mp\_int without modifying $a$ before 2256the quotient is obtained. 2257 2258The remainder of the source code is essentially the same as the source code for mp\_mul\_2d. The only significant difference is 2259the direction of the shifts. 2260 2261\subsection{Remainder of Division by Power of Two} 2262 2263The last algorithm in the series of polynomial basis power of two algorithms is calculating the remainder of division by $2^b$. This 2264algorithm benefits from the fact that in twos complement arithmetic $a \mbox{ (mod }2^b\mbox{)}$ is the same as $a$ AND $2^b - 1$. 2265 2266\begin{figure}[!here] 2267\begin{small} 2268\begin{center} 2269\begin{tabular}{l} 2270\hline Algorithm \textbf{mp\_mod\_2d}. \\ 2271\textbf{Input}. One mp\_int $a$ and an integer $b$ \\ 2272\textbf{Output}. $c \leftarrow a \mbox{ (mod }2^b\mbox{)}$. \\ 2273\hline \\ 22741. If $b \le 0$ then do \\ 2275\hspace{3mm}1.1 $c \leftarrow 0$ (\textit{mp\_zero}) \\ 2276\hspace{3mm}1.2 Return(\textit{MP\_OKAY}). \\ 22772. If $b > a.used \cdot lg(\beta)$ then do \\ 2278\hspace{3mm}2.1 $c \leftarrow a$ (\textit{mp\_copy}) \\ 2279\hspace{3mm}2.2 Return the result of step 2.1. \\ 22803. $c \leftarrow a$ \\ 22814. If step 3 failed return(\textit{MP\_MEM}). \\ 22825. for $n$ from $\lceil b / lg(\beta) \rceil$ to $c.used$ do \\ 2283\hspace{3mm}5.1 $c_n \leftarrow 0$ \\ 22846. $k \leftarrow b \mbox{ (mod }lg(\beta)\mbox{)}$ \\ 22857. $c_{\lfloor b / lg(\beta) \rfloor} \leftarrow c_{\lfloor b / lg(\beta) \rfloor} \mbox{ (mod }2^{k}\mbox{)}$. \\ 22868. Clamp excess digits of $c$. (\textit{mp\_clamp}) \\ 22879. Return(\textit{MP\_OKAY}). \\ 2288\hline 2289\end{tabular} 2290\end{center} 2291\end{small} 2292\caption{Algorithm mp\_mod\_2d} 2293\end{figure} 2294 2295\textbf{Algorithm mp\_mod\_2d.} 2296This algorithm will quickly calculate the value of $a \mbox{ (mod }2^b\mbox{)}$. First if $b$ is less than or equal to zero the 2297result is set to zero. If $b$ is greater than the number of bits in $a$ then it simply copies $a$ to $c$ and returns. Otherwise, $a$ 2298is copied to $b$, leading digits are removed and the remaining leading digit is trimed to the exact bit count. 2299 2300EXAM,bn_mp_mod_2d.c 2301 2302We first avoid cases of $b \le 0$ by simply mp\_zero()'ing the destination in such cases. Next if $2^b$ is larger 2303than the input we just mp\_copy() the input and return right away. After this point we know we must actually 2304perform some work to produce the remainder. 2305 2306Recalling that reducing modulo $2^k$ and a binary ``and'' with $2^k - 1$ are numerically equivalent we can quickly reduce 2307the number. First we zero any digits above the last digit in $2^b$ (line @41,for@). Next we reduce the 2308leading digit of both (line @45,&=@) and then mp\_clamp(). 2309 2310\section*{Exercises} 2311\begin{tabular}{cl} 2312$\left [ 3 \right ] $ & Devise an algorithm that performs $a \cdot 2^b$ for generic values of $b$ \\ 2313 & in $O(n)$ time. \\ 2314 &\\ 2315$\left [ 3 \right ] $ & Devise an efficient algorithm to multiply by small low hamming \\ 2316 & weight values such as $3$, $5$ and $9$. Extend it to handle all values \\ 2317 & upto $64$ with a hamming weight less than three. \\ 2318 &\\ 2319$\left [ 2 \right ] $ & Modify the preceding algorithm to handle values of the form \\ 2320 & $2^k - 1$ as well. \\ 2321 &\\ 2322$\left [ 3 \right ] $ & Using only algorithms mp\_mul\_2, mp\_div\_2 and mp\_add create an \\ 2323 & algorithm to multiply two integers in roughly $O(2n^2)$ time for \\ 2324 & any $n$-bit input. Note that the time of addition is ignored in the \\ 2325 & calculation. \\ 2326 & \\ 2327$\left [ 5 \right ] $ & Improve the previous algorithm to have a working time of at most \\ 2328 & $O \left (2^{(k-1)}n + \left ({2n^2 \over k} \right ) \right )$ for an appropriate choice of $k$. Again ignore \\ 2329 & the cost of addition. \\ 2330 & \\ 2331$\left [ 2 \right ] $ & Devise a chart to find optimal values of $k$ for the previous problem \\ 2332 & for $n = 64 \ldots 1024$ in steps of $64$. \\ 2333 & \\ 2334$\left [ 2 \right ] $ & Using only algorithms mp\_abs and mp\_sub devise another method for \\ 2335 & calculating the result of a signed comparison. \\ 2336 & 2337\end{tabular} 2338 2339\chapter{Multiplication and Squaring} 2340\section{The Multipliers} 2341For most number theoretic problems including certain public key cryptographic algorithms, the ``multipliers'' form the most important subset of 2342algorithms of any multiple precision integer package. The set of multiplier algorithms include integer multiplication, squaring and modular reduction 2343where in each of the algorithms single precision multiplication is the dominant operation performed. This chapter will discuss integer multiplication 2344and squaring, leaving modular reductions for the subsequent chapter. 2345 2346The importance of the multiplier algorithms is for the most part driven by the fact that certain popular public key algorithms are based on modular 2347exponentiation, that is computing $d \equiv a^b \mbox{ (mod }c\mbox{)}$ for some arbitrary choice of $a$, $b$, $c$ and $d$. During a modular 2348exponentiation the majority\footnote{Roughly speaking a modular exponentiation will spend about 40\% of the time performing modular reductions, 234935\% of the time performing squaring and 25\% of the time performing multiplications.} of the processor time is spent performing single precision 2350multiplications. 2351 2352For centuries general purpose multiplication has required a lengthly $O(n^2)$ process, whereby each digit of one multiplicand has to be multiplied 2353against every digit of the other multiplicand. Traditional long-hand multiplication is based on this process; while the techniques can differ the 2354overall algorithm used is essentially the same. Only ``recently'' have faster algorithms been studied. First Karatsuba multiplication was discovered in 23551962. This algorithm can multiply two numbers with considerably fewer single precision multiplications when compared to the long-hand approach. 2356This technique led to the discovery of polynomial basis algorithms (\textit{good reference?}) and subquently Fourier Transform based solutions. 2357 2358\section{Multiplication} 2359\subsection{The Baseline Multiplication} 2360\label{sec:basemult} 2361\index{baseline multiplication} 2362Computing the product of two integers in software can be achieved using a trivial adaptation of the standard $O(n^2)$ long-hand multiplication 2363algorithm that school children are taught. The algorithm is considered an $O(n^2)$ algorithm since for two $n$-digit inputs $n^2$ single precision 2364multiplications are required. More specifically for a $m$ and $n$ digit input $m \cdot n$ single precision multiplications are required. To 2365simplify most discussions, it will be assumed that the inputs have comparable number of digits. 2366 2367The ``baseline multiplication'' algorithm is designed to act as the ``catch-all'' algorithm, only to be used when the faster algorithms cannot be 2368used. This algorithm does not use any particularly interesting optimizations and should ideally be avoided if possible. One important 2369facet of this algorithm, is that it has been modified to only produce a certain amount of output digits as resolution. The importance of this 2370modification will become evident during the discussion of Barrett modular reduction. Recall that for a $n$ and $m$ digit input the product 2371will be at most $n + m$ digits. Therefore, this algorithm can be reduced to a full multiplier by having it produce $n + m$ digits of the product. 2372 2373Recall from ~GAMMA~ the definition of $\gamma$ as the number of bits in the type \textbf{mp\_digit}. We shall now extend the variable set to 2374include $\alpha$ which shall represent the number of bits in the type \textbf{mp\_word}. This implies that $2^{\alpha} > 2 \cdot \beta^2$. The 2375constant $\delta = 2^{\alpha - 2lg(\beta)}$ will represent the maximal weight of any column in a product (\textit{see ~COMBA~ for more information}). 2376 2377\newpage\begin{figure}[!here] 2378\begin{small} 2379\begin{center} 2380\begin{tabular}{l} 2381\hline Algorithm \textbf{s\_mp\_mul\_digs}. \\ 2382\textbf{Input}. mp\_int $a$, mp\_int $b$ and an integer $digs$ \\ 2383\textbf{Output}. $c \leftarrow \vert a \vert \cdot \vert b \vert \mbox{ (mod }\beta^{digs}\mbox{)}$. \\ 2384\hline \\ 23851. If min$(a.used, b.used) < \delta$ then do \\ 2386\hspace{3mm}1.1 Calculate $c = \vert a \vert \cdot \vert b \vert$ by the Comba method (\textit{see algorithm~\ref{fig:COMBAMULT}}). \\ 2387\hspace{3mm}1.2 Return the result of step 1.1 \\ 2388\\ 2389Allocate and initialize a temporary mp\_int. \\ 23902. Init $t$ to be of size $digs$ \\ 23913. If step 2 failed return(\textit{MP\_MEM}). \\ 23924. $t.used \leftarrow digs$ \\ 2393\\ 2394Compute the product. \\ 23955. for $ix$ from $0$ to $a.used - 1$ do \\ 2396\hspace{3mm}5.1 $u \leftarrow 0$ \\ 2397\hspace{3mm}5.2 $pb \leftarrow \mbox{min}(b.used, digs - ix)$ \\ 2398\hspace{3mm}5.3 If $pb < 1$ then goto step 6. \\ 2399\hspace{3mm}5.4 for $iy$ from $0$ to $pb - 1$ do \\ 2400\hspace{6mm}5.4.1 $\hat r \leftarrow t_{iy + ix} + a_{ix} \cdot b_{iy} + u$ \\ 2401\hspace{6mm}5.4.2 $t_{iy + ix} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\ 2402\hspace{6mm}5.4.3 $u \leftarrow \lfloor \hat r / \beta \rfloor$ \\ 2403\hspace{3mm}5.5 if $ix + pb < digs$ then do \\ 2404\hspace{6mm}5.5.1 $t_{ix + pb} \leftarrow u$ \\ 24056. Clamp excess digits of $t$. \\ 24067. Swap $c$ with $t$ \\ 24078. Clear $t$ \\ 24089. Return(\textit{MP\_OKAY}). \\ 2409\hline 2410\end{tabular} 2411\end{center} 2412\end{small} 2413\caption{Algorithm s\_mp\_mul\_digs} 2414\end{figure} 2415 2416\textbf{Algorithm s\_mp\_mul\_digs.} 2417This algorithm computes the unsigned product of two inputs $a$ and $b$, limited to an output precision of $digs$ digits. While it may seem 2418a bit awkward to modify the function from its simple $O(n^2)$ description, the usefulness of partial multipliers will arise in a subsequent 2419algorithm. The algorithm is loosely based on algorithm 14.12 from \cite[pp. 595]{HAC} and is similar to Algorithm M of Knuth \cite[pp. 268]{TAOCPV2}. 2420Algorithm s\_mp\_mul\_digs differs from these cited references since it can produce a variable output precision regardless of the precision of the 2421inputs. 2422 2423The first thing this algorithm checks for is whether a Comba multiplier can be used instead. If the minimum digit count of either 2424input is less than $\delta$, then the Comba method may be used instead. After the Comba method is ruled out, the baseline algorithm begins. A 2425temporary mp\_int variable $t$ is used to hold the intermediate result of the product. This allows the algorithm to be used to 2426compute products when either $a = c$ or $b = c$ without overwriting the inputs. 2427 2428All of step 5 is the infamous $O(n^2)$ multiplication loop slightly modified to only produce upto $digs$ digits of output. The $pb$ variable 2429is given the count of digits to read from $b$ inside the nested loop. If $pb \le 1$ then no more output digits can be produced and the algorithm 2430will exit the loop. The best way to think of the loops are as a series of $pb \times 1$ multiplications. That is, in each pass of the 2431innermost loop $a_{ix}$ is multiplied against $b$ and the result is added (\textit{with an appropriate shift}) to $t$. 2432 2433For example, consider multiplying $576$ by $241$. That is equivalent to computing $10^0(1)(576) + 10^1(4)(576) + 10^2(2)(576)$ which is best 2434visualized in the following table. 2435 2436\begin{figure}[here] 2437\begin{center} 2438\begin{tabular}{|c|c|c|c|c|c|l|} 2439\hline && & 5 & 7 & 6 & \\ 2440\hline $\times$&& & 2 & 4 & 1 & \\ 2441\hline &&&&&&\\ 2442 && & 5 & 7 & 6 & $10^0(1)(576)$ \\ 2443 &2 & 3 & 6 & 1 & 6 & $10^1(4)(576) + 10^0(1)(576)$ \\ 2444 1 & 3 & 8 & 8 & 1 & 6 & $10^2(2)(576) + 10^1(4)(576) + 10^0(1)(576)$ \\ 2445\hline 2446\end{tabular} 2447\end{center} 2448\caption{Long-Hand Multiplication Diagram} 2449\end{figure} 2450 2451Each row of the product is added to the result after being shifted to the left (\textit{multiplied by a power of the radix}) by the appropriate 2452count. That is in pass $ix$ of the inner loop the product is added starting at the $ix$'th digit of the reult. 2453 2454Step 5.4.1 introduces the hat symbol (\textit{e.g. $\hat r$}) which represents a double precision variable. The multiplication on that step 2455is assumed to be a double wide output single precision multiplication. That is, two single precision variables are multiplied to produce a 2456double precision result. The step is somewhat optimized from a long-hand multiplication algorithm because the carry from the addition in step 24575.4.1 is propagated through the nested loop. If the carry was not propagated immediately it would overflow the single precision digit 2458$t_{ix+iy}$ and the result would be lost. 2459 2460At step 5.5 the nested loop is finished and any carry that was left over should be forwarded. The carry does not have to be added to the $ix+pb$'th 2461digit since that digit is assumed to be zero at this point. However, if $ix + pb \ge digs$ the carry is not set as it would make the result 2462exceed the precision requested. 2463 2464EXAM,bn_s_mp_mul_digs.c 2465 2466First we determine (line @30,if@) if the Comba method can be used first since it's faster. The conditions for 2467sing the Comba routine are that min$(a.used, b.used) < \delta$ and the number of digits of output is less than 2468\textbf{MP\_WARRAY}. This new constant is used to control the stack usage in the Comba routines. By default it is 2469set to $\delta$ but can be reduced when memory is at a premium. 2470 2471If we cannot use the Comba method we proceed to setup the baseline routine. We allocate the the destination mp\_int 2472$t$ (line @36,init@) to the exact size of the output to avoid further re--allocations. At this point we now 2473begin the $O(n^2)$ loop. 2474 2475This implementation of multiplication has the caveat that it can be trimmed to only produce a variable number of 2476digits as output. In each iteration of the outer loop the $pb$ variable is set (line @48,MIN@) to the maximum 2477number of inner loop iterations. 2478 2479Inside the inner loop we calculate $\hat r$ as the mp\_word product of the two mp\_digits and the addition of the 2480carry from the previous iteration. A particularly important observation is that most modern optimizing 2481C compilers (GCC for instance) can recognize that a $N \times N \rightarrow 2N$ multiplication is all that 2482is required for the product. In x86 terms for example, this means using the MUL instruction. 2483 2484Each digit of the product is stored in turn (line @68,tmpt@) and the carry propagated (line @71,>>@) to the 2485next iteration. 2486 2487\subsection{Faster Multiplication by the ``Comba'' Method} 2488MARK,COMBA 2489 2490One of the huge drawbacks of the ``baseline'' algorithms is that at the $O(n^2)$ level the carry must be 2491computed and propagated upwards. This makes the nested loop very sequential and hard to unroll and implement 2492in parallel. The ``Comba'' \cite{COMBA} method is named after little known (\textit{in cryptographic venues}) Paul G. 2493Comba who described a method of implementing fast multipliers that do not require nested carry fixup operations. As an 2494interesting aside it seems that Paul Barrett describes a similar technique in his 1986 paper \cite{BARRETT} written 2495five years before. 2496 2497At the heart of the Comba technique is once again the long-hand algorithm. Except in this case a slight 2498twist is placed on how the columns of the result are produced. In the standard long-hand algorithm rows of products 2499are produced then added together to form the final result. In the baseline algorithm the columns are added together 2500after each iteration to get the result instantaneously. 2501 2502In the Comba algorithm the columns of the result are produced entirely independently of each other. That is at 2503the $O(n^2)$ level a simple multiplication and addition step is performed. The carries of the columns are propagated 2504after the nested loop to reduce the amount of work requiored. Succintly the first step of the algorithm is to compute 2505the product vector $\vec x$ as follows. 2506 2507\begin{equation} 2508\vec x_n = \sum_{i+j = n} a_ib_j, \forall n \in \lbrace 0, 1, 2, \ldots, i + j \rbrace 2509\end{equation} 2510 2511Where $\vec x_n$ is the $n'th$ column of the output vector. Consider the following example which computes the vector $\vec x$ for the multiplication 2512of $576$ and $241$. 2513 2514\newpage\begin{figure}[here] 2515\begin{small} 2516\begin{center} 2517\begin{tabular}{|c|c|c|c|c|c|} 2518 \hline & & 5 & 7 & 6 & First Input\\ 2519 \hline $\times$ & & 2 & 4 & 1 & Second Input\\ 2520\hline & & $1 \cdot 5 = 5$ & $1 \cdot 7 = 7$ & $1 \cdot 6 = 6$ & First pass \\ 2521 & $4 \cdot 5 = 20$ & $4 \cdot 7+5=33$ & $4 \cdot 6+7=31$ & 6 & Second pass \\ 2522 $2 \cdot 5 = 10$ & $2 \cdot 7 + 20 = 34$ & $2 \cdot 6+33=45$ & 31 & 6 & Third pass \\ 2523\hline 10 & 34 & 45 & 31 & 6 & Final Result \\ 2524\hline 2525\end{tabular} 2526\end{center} 2527\end{small} 2528\caption{Comba Multiplication Diagram} 2529\end{figure} 2530 2531At this point the vector $x = \left < 10, 34, 45, 31, 6 \right >$ is the result of the first step of the Comba multipler. 2532Now the columns must be fixed by propagating the carry upwards. The resultant vector will have one extra dimension over the input vector which is 2533congruent to adding a leading zero digit. 2534 2535\begin{figure}[!here] 2536\begin{small} 2537\begin{center} 2538\begin{tabular}{l} 2539\hline Algorithm \textbf{Comba Fixup}. \\ 2540\textbf{Input}. Vector $\vec x$ of dimension $k$ \\ 2541\textbf{Output}. Vector $\vec x$ such that the carries have been propagated. \\ 2542\hline \\ 25431. for $n$ from $0$ to $k - 1$ do \\ 2544\hspace{3mm}1.1 $\vec x_{n+1} \leftarrow \vec x_{n+1} + \lfloor \vec x_{n}/\beta \rfloor$ \\ 2545\hspace{3mm}1.2 $\vec x_{n} \leftarrow \vec x_{n} \mbox{ (mod }\beta\mbox{)}$ \\ 25462. Return($\vec x$). \\ 2547\hline 2548\end{tabular} 2549\end{center} 2550\end{small} 2551\caption{Algorithm Comba Fixup} 2552\end{figure} 2553 2554With that algorithm and $k = 5$ and $\beta = 10$ the following vector is produced $\vec x= \left < 1, 3, 8, 8, 1, 6 \right >$. In this case 2555$241 \cdot 576$ is in fact $138816$ and the procedure succeeded. If the algorithm is correct and as will be demonstrated shortly more 2556efficient than the baseline algorithm why not simply always use this algorithm? 2557 2558\subsubsection{Column Weight.} 2559At the nested $O(n^2)$ level the Comba method adds the product of two single precision variables to each column of the output 2560independently. A serious obstacle is if the carry is lost, due to lack of precision before the algorithm has a chance to fix 2561the carries. For example, in the multiplication of two three-digit numbers the third column of output will be the sum of 2562three single precision multiplications. If the precision of the accumulator for the output digits is less then $3 \cdot (\beta - 1)^2$ then 2563an overflow can occur and the carry information will be lost. For any $m$ and $n$ digit inputs the maximum weight of any column is 2564min$(m, n)$ which is fairly obvious. 2565 2566The maximum number of terms in any column of a product is known as the ``column weight'' and strictly governs when the algorithm can be used. Recall 2567from earlier that a double precision type has $\alpha$ bits of resolution and a single precision digit has $lg(\beta)$ bits of precision. Given these 2568two quantities we must not violate the following 2569 2570\begin{equation} 2571k \cdot \left (\beta - 1 \right )^2 < 2^{\alpha} 2572\end{equation} 2573 2574Which reduces to 2575 2576\begin{equation} 2577k \cdot \left ( \beta^2 - 2\beta + 1 \right ) < 2^{\alpha} 2578\end{equation} 2579 2580Let $\rho = lg(\beta)$ represent the number of bits in a single precision digit. By further re-arrangement of the equation the final solution is 2581found. 2582 2583\begin{equation} 2584k < {{2^{\alpha}} \over {\left (2^{2\rho} - 2^{\rho + 1} + 1 \right )}} 2585\end{equation} 2586 2587The defaults for LibTomMath are $\beta = 2^{28}$ and $\alpha = 2^{64}$ which means that $k$ is bounded by $k < 257$. In this configuration 2588the smaller input may not have more than $256$ digits if the Comba method is to be used. This is quite satisfactory for most applications since 2589$256$ digits would allow for numbers in the range of $0 \le x < 2^{7168}$ which, is much larger than most public key cryptographic algorithms require. 2590 2591\newpage\begin{figure}[!here] 2592\begin{small} 2593\begin{center} 2594\begin{tabular}{l} 2595\hline Algorithm \textbf{fast\_s\_mp\_mul\_digs}. \\ 2596\textbf{Input}. mp\_int $a$, mp\_int $b$ and an integer $digs$ \\ 2597\textbf{Output}. $c \leftarrow \vert a \vert \cdot \vert b \vert \mbox{ (mod }\beta^{digs}\mbox{)}$. \\ 2598\hline \\ 2599Place an array of \textbf{MP\_WARRAY} single precision digits named $W$ on the stack. \\ 26001. If $c.alloc < digs$ then grow $c$ to $digs$ digits. (\textit{mp\_grow}) \\ 26012. If step 1 failed return(\textit{MP\_MEM}).\\ 2602\\ 26033. $pa \leftarrow \mbox{MIN}(digs, a.used + b.used)$ \\ 2604\\ 26054. $\_ \hat W \leftarrow 0$ \\ 26065. for $ix$ from 0 to $pa - 1$ do \\ 2607\hspace{3mm}5.1 $ty \leftarrow \mbox{MIN}(b.used - 1, ix)$ \\ 2608\hspace{3mm}5.2 $tx \leftarrow ix - ty$ \\ 2609\hspace{3mm}5.3 $iy \leftarrow \mbox{MIN}(a.used - tx, ty + 1)$ \\ 2610\hspace{3mm}5.4 for $iz$ from 0 to $iy - 1$ do \\ 2611\hspace{6mm}5.4.1 $\_ \hat W \leftarrow \_ \hat W + a_{tx+iy}b_{ty-iy}$ \\ 2612\hspace{3mm}5.5 $W_{ix} \leftarrow \_ \hat W (\mbox{mod }\beta)$\\ 2613\hspace{3mm}5.6 $\_ \hat W \leftarrow \lfloor \_ \hat W / \beta \rfloor$ \\ 2614\\ 26156. $oldused \leftarrow c.used$ \\ 26167. $c.used \leftarrow digs$ \\ 26178. for $ix$ from $0$ to $pa$ do \\ 2618\hspace{3mm}8.1 $c_{ix} \leftarrow W_{ix}$ \\ 26199. for $ix$ from $pa + 1$ to $oldused - 1$ do \\ 2620\hspace{3mm}9.1 $c_{ix} \leftarrow 0$ \\ 2621\\ 262210. Clamp $c$. \\ 262311. Return MP\_OKAY. \\ 2624\hline 2625\end{tabular} 2626\end{center} 2627\end{small} 2628\caption{Algorithm fast\_s\_mp\_mul\_digs} 2629\label{fig:COMBAMULT} 2630\end{figure} 2631 2632\textbf{Algorithm fast\_s\_mp\_mul\_digs.} 2633This algorithm performs the unsigned multiplication of $a$ and $b$ using the Comba method limited to $digs$ digits of precision. 2634 2635The outer loop of this algorithm is more complicated than that of the baseline multiplier. This is because on the inside of the 2636loop we want to produce one column per pass. This allows the accumulator $\_ \hat W$ to be placed in CPU registers and 2637reduce the memory bandwidth to two \textbf{mp\_digit} reads per iteration. 2638 2639The $ty$ variable is set to the minimum count of $ix$ or the number of digits in $b$. That way if $a$ has more digits than 2640$b$ this will be limited to $b.used - 1$. The $tx$ variable is set to the to the distance past $b.used$ the variable 2641$ix$ is. This is used for the immediately subsequent statement where we find $iy$. 2642 2643The variable $iy$ is the minimum digits we can read from either $a$ or $b$ before running out. Computing one column at a time 2644means we have to scan one integer upwards and the other downwards. $a$ starts at $tx$ and $b$ starts at $ty$. In each 2645pass we are producing the $ix$'th output column and we note that $tx + ty = ix$. As we move $tx$ upwards we have to 2646move $ty$ downards so the equality remains valid. The $iy$ variable is the number of iterations until 2647$tx \ge a.used$ or $ty < 0$ occurs. 2648 2649After every inner pass we store the lower half of the accumulator into $W_{ix}$ and then propagate the carry of the accumulator 2650into the next round by dividing $\_ \hat W$ by $\beta$. 2651 2652To measure the benefits of the Comba method over the baseline method consider the number of operations that are required. If the 2653cost in terms of time of a multiply and addition is $p$ and the cost of a carry propagation is $q$ then a baseline multiplication would require 2654$O \left ((p + q)n^2 \right )$ time to multiply two $n$-digit numbers. The Comba method requires only $O(pn^2 + qn)$ time, however in practice, 2655the speed increase is actually much more. With $O(n)$ space the algorithm can be reduced to $O(pn + qn)$ time by implementing the $n$ multiply 2656and addition operations in the nested loop in parallel. 2657 2658EXAM,bn_fast_s_mp_mul_digs.c 2659 2660As per the pseudo--code we first calculate $pa$ (line @47,MIN@) as the number of digits to output. Next we begin the outer loop 2661to produce the individual columns of the product. We use the two aliases $tmpx$ and $tmpy$ (lines @61,tmpx@, @62,tmpy@) to point 2662inside the two multiplicands quickly. 2663 2664The inner loop (lines @70,for@ to @72,}@) of this implementation is where the tradeoff come into play. Originally this comba 2665implementation was ``row--major'' which means it adds to each of the columns in each pass. After the outer loop it would then fix 2666the carries. This was very fast except it had an annoying drawback. You had to read a mp\_word and two mp\_digits and write 2667one mp\_word per iteration. On processors such as the Athlon XP and P4 this did not matter much since the cache bandwidth 2668is very high and it can keep the ALU fed with data. It did, however, matter on older and embedded cpus where cache is often 2669slower and also often doesn't exist. This new algorithm only performs two reads per iteration under the assumption that the 2670compiler has aliased $\_ \hat W$ to a CPU register. 2671 2672After the inner loop we store the current accumulator in $W$ and shift $\_ \hat W$ (lines @75,W[ix]@, @78,>>@) to forward it as 2673a carry for the next pass. After the outer loop we use the final carry (line @82,W[ix]@) as the last digit of the product. 2674 2675\subsection{Polynomial Basis Multiplication} 2676To break the $O(n^2)$ barrier in multiplication requires a completely different look at integer multiplication. In the following algorithms 2677the use of polynomial basis representation for two integers $a$ and $b$ as $f(x) = \sum_{i=0}^{n} a_i x^i$ and 2678$g(x) = \sum_{i=0}^{n} b_i x^i$ respectively, is required. In this system both $f(x)$ and $g(x)$ have $n + 1$ terms and are of the $n$'th degree. 2679 2680The product $a \cdot b \equiv f(x)g(x)$ is the polynomial $W(x) = \sum_{i=0}^{2n} w_i x^i$. The coefficients $w_i$ will 2681directly yield the desired product when $\beta$ is substituted for $x$. The direct solution to solve for the $2n + 1$ coefficients 2682requires $O(n^2)$ time and would in practice be slower than the Comba technique. 2683 2684However, numerical analysis theory indicates that only $2n + 1$ distinct points in $W(x)$ are required to determine the values of the $2n + 1$ unknown 2685coefficients. This means by finding $\zeta_y = W(y)$ for $2n + 1$ small values of $y$ the coefficients of $W(x)$ can be found with 2686Gaussian elimination. This technique is also occasionally refered to as the \textit{interpolation technique} (\textit{references please...}) since in 2687effect an interpolation based on $2n + 1$ points will yield a polynomial equivalent to $W(x)$. 2688 2689The coefficients of the polynomial $W(x)$ are unknown which makes finding $W(y)$ for any value of $y$ impossible. However, since 2690$W(x) = f(x)g(x)$ the equivalent $\zeta_y = f(y) g(y)$ can be used in its place. The benefit of this technique stems from the 2691fact that $f(y)$ and $g(y)$ are much smaller than either $a$ or $b$ respectively. As a result finding the $2n + 1$ relations required 2692by multiplying $f(y)g(y)$ involves multiplying integers that are much smaller than either of the inputs. 2693 2694When picking points to gather relations there are always three obvious points to choose, $y = 0, 1$ and $ \infty$. The $\zeta_0$ term 2695is simply the product $W(0) = w_0 = a_0 \cdot b_0$. The $\zeta_1$ term is the product 2696$W(1) = \left (\sum_{i = 0}^{n} a_i \right ) \left (\sum_{i = 0}^{n} b_i \right )$. The third point $\zeta_{\infty}$ is less obvious but rather 2697simple to explain. The $2n + 1$'th coefficient of $W(x)$ is numerically equivalent to the most significant column in an integer multiplication. 2698The point at $\infty$ is used symbolically to represent the most significant column, that is $W(\infty) = w_{2n} = a_nb_n$. Note that the 2699points at $y = 0$ and $\infty$ yield the coefficients $w_0$ and $w_{2n}$ directly. 2700 2701If more points are required they should be of small values and powers of two such as $2^q$ and the related \textit{mirror points} 2702$\left (2^q \right )^{2n} \cdot \zeta_{2^{-q}}$ for small values of $q$. The term ``mirror point'' stems from the fact that 2703$\left (2^q \right )^{2n} \cdot \zeta_{2^{-q}}$ can be calculated in the exact opposite fashion as $\zeta_{2^q}$. For 2704example, when $n = 2$ and $q = 1$ then following two equations are equivalent to the point $\zeta_{2}$ and its mirror. 2705 2706\begin{eqnarray} 2707\zeta_{2} = f(2)g(2) = (4a_2 + 2a_1 + a_0)(4b_2 + 2b_1 + b_0) \nonumber \\ 270816 \cdot \zeta_{1 \over 2} = 4f({1\over 2}) \cdot 4g({1 \over 2}) = (a_2 + 2a_1 + 4a_0)(b_2 + 2b_1 + 4b_0) 2709\end{eqnarray} 2710 2711Using such points will allow the values of $f(y)$ and $g(y)$ to be independently calculated using only left shifts. For example, when $n = 2$ the 2712polynomial $f(2^q)$ is equal to $2^q((2^qa_2) + a_1) + a_0$. This technique of polynomial representation is known as Horner's method. 2713 2714As a general rule of the algorithm when the inputs are split into $n$ parts each there are $2n - 1$ multiplications. Each multiplication is of 2715multiplicands that have $n$ times fewer digits than the inputs. The asymptotic running time of this algorithm is 2716$O \left ( k^{lg_n(2n - 1)} \right )$ for $k$ digit inputs (\textit{assuming they have the same number of digits}). Figure~\ref{fig:exponent} 2717summarizes the exponents for various values of $n$. 2718 2719\begin{figure} 2720\begin{center} 2721\begin{tabular}{|c|c|c|} 2722\hline \textbf{Split into $n$ Parts} & \textbf{Exponent} & \textbf{Notes}\\ 2723\hline $2$ & $1.584962501$ & This is Karatsuba Multiplication. \\ 2724\hline $3$ & $1.464973520$ & This is Toom-Cook Multiplication. \\ 2725\hline $4$ & $1.403677461$ &\\ 2726\hline $5$ & $1.365212389$ &\\ 2727\hline $10$ & $1.278753601$ &\\ 2728\hline $100$ & $1.149426538$ &\\ 2729\hline $1000$ & $1.100270931$ &\\ 2730\hline $10000$ & $1.075252070$ &\\ 2731\hline 2732\end{tabular} 2733\end{center} 2734\caption{Asymptotic Running Time of Polynomial Basis Multiplication} 2735\label{fig:exponent} 2736\end{figure} 2737 2738At first it may seem like a good idea to choose $n = 1000$ since the exponent is approximately $1.1$. However, the overhead 2739of solving for the 2001 terms of $W(x)$ will certainly consume any savings the algorithm could offer for all but exceedingly large 2740numbers. 2741 2742\subsubsection{Cutoff Point} 2743The polynomial basis multiplication algorithms all require fewer single precision multiplications than a straight Comba approach. However, 2744the algorithms incur an overhead (\textit{at the $O(n)$ work level}) since they require a system of equations to be solved. This makes the 2745polynomial basis approach more costly to use with small inputs. 2746 2747Let $m$ represent the number of digits in the multiplicands (\textit{assume both multiplicands have the same number of digits}). There exists a 2748point $y$ such that when $m < y$ the polynomial basis algorithms are more costly than Comba, when $m = y$ they are roughly the same cost and 2749when $m > y$ the Comba methods are slower than the polynomial basis algorithms. 2750 2751The exact location of $y$ depends on several key architectural elements of the computer platform in question. 2752 2753\begin{enumerate} 2754\item The ratio of clock cycles for single precision multiplication versus other simpler operations such as addition, shifting, etc. For example 2755on the AMD Athlon the ratio is roughly $17 : 1$ while on the Intel P4 it is $29 : 1$. The higher the ratio in favour of multiplication the lower 2756the cutoff point $y$ will be. 2757 2758\item The complexity of the linear system of equations (\textit{for the coefficients of $W(x)$}) is. Generally speaking as the number of splits 2759grows the complexity grows substantially. Ideally solving the system will only involve addition, subtraction and shifting of integers. This 2760directly reflects on the ratio previous mentioned. 2761 2762\item To a lesser extent memory bandwidth and function call overheads. Provided the values are in the processor cache this is less of an 2763influence over the cutoff point. 2764 2765\end{enumerate} 2766 2767A clean cutoff point separation occurs when a point $y$ is found such that all of the cutoff point conditions are met. For example, if the point 2768is too low then there will be values of $m$ such that $m > y$ and the Comba method is still faster. Finding the cutoff points is fairly simple when 2769a high resolution timer is available. 2770 2771\subsection{Karatsuba Multiplication} 2772Karatsuba \cite{KARA} multiplication when originally proposed in 1962 was among the first set of algorithms to break the $O(n^2)$ barrier for 2773general purpose multiplication. Given two polynomial basis representations $f(x) = ax + b$ and $g(x) = cx + d$, Karatsuba proved with 2774light algebra \cite{KARAP} that the following polynomial is equivalent to multiplication of the two integers the polynomials represent. 2775 2776\begin{equation} 2777f(x) \cdot g(x) = acx^2 + ((a + b)(c + d) - (ac + bd))x + bd 2778\end{equation} 2779 2780Using the observation that $ac$ and $bd$ could be re-used only three half sized multiplications would be required to produce the product. Applying 2781this algorithm recursively, the work factor becomes $O(n^{lg(3)})$ which is substantially better than the work factor $O(n^2)$ of the Comba technique. It turns 2782out what Karatsuba did not know or at least did not publish was that this is simply polynomial basis multiplication with the points 2783$\zeta_0$, $\zeta_{\infty}$ and $\zeta_{1}$. Consider the resultant system of equations. 2784 2785\begin{center} 2786\begin{tabular}{rcrcrcrc} 2787$\zeta_{0}$ & $=$ & & & & & $w_0$ \\ 2788$\zeta_{1}$ & $=$ & $w_2$ & $+$ & $w_1$ & $+$ & $w_0$ \\ 2789$\zeta_{\infty}$ & $=$ & $w_2$ & & & & \\ 2790\end{tabular} 2791\end{center} 2792 2793By adding the first and last equation to the equation in the middle the term $w_1$ can be isolated and all three coefficients solved for. The simplicity 2794of this system of equations has made Karatsuba fairly popular. In fact the cutoff point is often fairly low\footnote{With LibTomMath 0.18 it is 70 and 109 digits for the Intel P4 and AMD Athlon respectively.} 2795making it an ideal algorithm to speed up certain public key cryptosystems such as RSA and Diffie-Hellman. 2796 2797\newpage\begin{figure}[!here] 2798\begin{small} 2799\begin{center} 2800\begin{tabular}{l} 2801\hline Algorithm \textbf{mp\_karatsuba\_mul}. \\ 2802\textbf{Input}. mp\_int $a$ and mp\_int $b$ \\ 2803\textbf{Output}. $c \leftarrow \vert a \vert \cdot \vert b \vert$ \\ 2804\hline \\ 28051. Init the following mp\_int variables: $x0$, $x1$, $y0$, $y1$, $t1$, $x0y0$, $x1y1$.\\ 28062. If step 2 failed then return(\textit{MP\_MEM}). \\ 2807\\ 2808Split the input. e.g. $a = x1 \cdot \beta^B + x0$ \\ 28093. $B \leftarrow \mbox{min}(a.used, b.used)/2$ \\ 28104. $x0 \leftarrow a \mbox{ (mod }\beta^B\mbox{)}$ (\textit{mp\_mod\_2d}) \\ 28115. $y0 \leftarrow b \mbox{ (mod }\beta^B\mbox{)}$ \\ 28126. $x1 \leftarrow \lfloor a / \beta^B \rfloor$ (\textit{mp\_rshd}) \\ 28137. $y1 \leftarrow \lfloor b / \beta^B \rfloor$ \\ 2814\\ 2815Calculate the three products. \\ 28168. $x0y0 \leftarrow x0 \cdot y0$ (\textit{mp\_mul}) \\ 28179. $x1y1 \leftarrow x1 \cdot y1$ \\ 281810. $t1 \leftarrow x1 + x0$ (\textit{mp\_add}) \\ 281911. $x0 \leftarrow y1 + y0$ \\ 282012. $t1 \leftarrow t1 \cdot x0$ \\ 2821\\ 2822Calculate the middle term. \\ 282313. $x0 \leftarrow x0y0 + x1y1$ \\ 282414. $t1 \leftarrow t1 - x0$ (\textit{s\_mp\_sub}) \\ 2825\\ 2826Calculate the final product. \\ 282715. $t1 \leftarrow t1 \cdot \beta^B$ (\textit{mp\_lshd}) \\ 282816. $x1y1 \leftarrow x1y1 \cdot \beta^{2B}$ \\ 282917. $t1 \leftarrow x0y0 + t1$ \\ 283018. $c \leftarrow t1 + x1y1$ \\ 283119. Clear all of the temporary variables. \\ 283220. Return(\textit{MP\_OKAY}).\\ 2833\hline 2834\end{tabular} 2835\end{center} 2836\end{small} 2837\caption{Algorithm mp\_karatsuba\_mul} 2838\end{figure} 2839 2840\textbf{Algorithm mp\_karatsuba\_mul.} 2841This algorithm computes the unsigned product of two inputs using the Karatsuba multiplication algorithm. It is loosely based on the description 2842from Knuth \cite[pp. 294-295]{TAOCPV2}. 2843 2844\index{radix point} 2845In order to split the two inputs into their respective halves, a suitable \textit{radix point} must be chosen. The radix point chosen must 2846be used for both of the inputs meaning that it must be smaller than the smallest input. Step 3 chooses the radix point $B$ as half of the 2847smallest input \textbf{used} count. After the radix point is chosen the inputs are split into lower and upper halves. Step 4 and 5 2848compute the lower halves. Step 6 and 7 computer the upper halves. 2849 2850After the halves have been computed the three intermediate half-size products must be computed. Step 8 and 9 compute the trivial products 2851$x0 \cdot y0$ and $x1 \cdot y1$. The mp\_int $x0$ is used as a temporary variable after $x1 + x0$ has been computed. By using $x0$ instead 2852of an additional temporary variable, the algorithm can avoid an addition memory allocation operation. 2853 2854The remaining steps 13 through 18 compute the Karatsuba polynomial through a variety of digit shifting and addition operations. 2855 2856EXAM,bn_mp_karatsuba_mul.c 2857 2858The new coding element in this routine, not seen in previous routines, is the usage of goto statements. The conventional 2859wisdom is that goto statements should be avoided. This is generally true, however when every single function call can fail, it makes sense 2860to handle error recovery with a single piece of code. Lines @61,if@ to @75,if@ handle initializing all of the temporary variables 2861required. Note how each of the if statements goes to a different label in case of failure. This allows the routine to correctly free only 2862the temporaries that have been successfully allocated so far. 2863 2864The temporary variables are all initialized using the mp\_init\_size routine since they are expected to be large. This saves the 2865additional reallocation that would have been necessary. Also $x0$, $x1$, $y0$ and $y1$ have to be able to hold at least their respective 2866number of digits for the next section of code. 2867 2868The first algebraic portion of the algorithm is to split the two inputs into their halves. However, instead of using mp\_mod\_2d and mp\_rshd 2869to extract the halves, the respective code has been placed inline within the body of the function. To initialize the halves, the \textbf{used} and 2870\textbf{sign} members are copied first. The first for loop on line @98,for@ copies the lower halves. Since they are both the same magnitude it 2871is simpler to calculate both lower halves in a single loop. The for loop on lines @104,for@ and @109,for@ calculate the upper halves $x1$ and 2872$y1$ respectively. 2873 2874By inlining the calculation of the halves, the Karatsuba multiplier has a slightly lower overhead and can be used for smaller magnitude inputs. 2875 2876When line @152,err@ is reached, the algorithm has completed succesfully. The ``error status'' variable $err$ is set to \textbf{MP\_OKAY} so that 2877the same code that handles errors can be used to clear the temporary variables and return. 2878 2879\subsection{Toom-Cook $3$-Way Multiplication} 2880Toom-Cook $3$-Way \cite{TOOM} multiplication is essentially the polynomial basis algorithm for $n = 2$ except that the points are 2881chosen such that $\zeta$ is easy to compute and the resulting system of equations easy to reduce. Here, the points $\zeta_{0}$, 2882$16 \cdot \zeta_{1 \over 2}$, $\zeta_1$, $\zeta_2$ and $\zeta_{\infty}$ make up the five required points to solve for the coefficients 2883of the $W(x)$. 2884 2885With the five relations that Toom-Cook specifies, the following system of equations is formed. 2886 2887\begin{center} 2888\begin{tabular}{rcrcrcrcrcr} 2889$\zeta_0$ & $=$ & $0w_4$ & $+$ & $0w_3$ & $+$ & $0w_2$ & $+$ & $0w_1$ & $+$ & $1w_0$ \\ 2890$16 \cdot \zeta_{1 \over 2}$ & $=$ & $1w_4$ & $+$ & $2w_3$ & $+$ & $4w_2$ & $+$ & $8w_1$ & $+$ & $16w_0$ \\ 2891$\zeta_1$ & $=$ & $1w_4$ & $+$ & $1w_3$ & $+$ & $1w_2$ & $+$ & $1w_1$ & $+$ & $1w_0$ \\ 2892$\zeta_2$ & $=$ & $16w_4$ & $+$ & $8w_3$ & $+$ & $4w_2$ & $+$ & $2w_1$ & $+$ & $1w_0$ \\ 2893$\zeta_{\infty}$ & $=$ & $1w_4$ & $+$ & $0w_3$ & $+$ & $0w_2$ & $+$ & $0w_1$ & $+$ & $0w_0$ \\ 2894\end{tabular} 2895\end{center} 2896 2897A trivial solution to this matrix requires $12$ subtractions, two multiplications by a small power of two, two divisions by a small power 2898of two, two divisions by three and one multiplication by three. All of these $19$ sub-operations require less than quadratic time, meaning that 2899the algorithm can be faster than a baseline multiplication. However, the greater complexity of this algorithm places the cutoff point 2900(\textbf{TOOM\_MUL\_CUTOFF}) where Toom-Cook becomes more efficient much higher than the Karatsuba cutoff point. 2901 2902\begin{figure}[!here] 2903\begin{small} 2904\begin{center} 2905\begin{tabular}{l} 2906\hline Algorithm \textbf{mp\_toom\_mul}. \\ 2907\textbf{Input}. mp\_int $a$ and mp\_int $b$ \\ 2908\textbf{Output}. $c \leftarrow a \cdot b $ \\ 2909\hline \\ 2910Split $a$ and $b$ into three pieces. E.g. $a = a_2 \beta^{2k} + a_1 \beta^{k} + a_0$ \\ 29111. $k \leftarrow \lfloor \mbox{min}(a.used, b.used) / 3 \rfloor$ \\ 29122. $a_0 \leftarrow a \mbox{ (mod }\beta^{k}\mbox{)}$ \\ 29133. $a_1 \leftarrow \lfloor a / \beta^k \rfloor$, $a_1 \leftarrow a_1 \mbox{ (mod }\beta^{k}\mbox{)}$ \\ 29144. $a_2 \leftarrow \lfloor a / \beta^{2k} \rfloor$, $a_2 \leftarrow a_2 \mbox{ (mod }\beta^{k}\mbox{)}$ \\ 29155. $b_0 \leftarrow a \mbox{ (mod }\beta^{k}\mbox{)}$ \\ 29166. $b_1 \leftarrow \lfloor a / \beta^k \rfloor$, $b_1 \leftarrow b_1 \mbox{ (mod }\beta^{k}\mbox{)}$ \\ 29177. $b_2 \leftarrow \lfloor a / \beta^{2k} \rfloor$, $b_2 \leftarrow b_2 \mbox{ (mod }\beta^{k}\mbox{)}$ \\ 2918\\ 2919Find the five equations for $w_0, w_1, ..., w_4$. \\ 29208. $w_0 \leftarrow a_0 \cdot b_0$ \\ 29219. $w_4 \leftarrow a_2 \cdot b_2$ \\ 292210. $tmp_1 \leftarrow 2 \cdot a_0$, $tmp_1 \leftarrow a_1 + tmp_1$, $tmp_1 \leftarrow 2 \cdot tmp_1$, $tmp_1 \leftarrow tmp_1 + a_2$ \\ 292311. $tmp_2 \leftarrow 2 \cdot b_0$, $tmp_2 \leftarrow b_1 + tmp_2$, $tmp_2 \leftarrow 2 \cdot tmp_2$, $tmp_2 \leftarrow tmp_2 + b_2$ \\ 292412. $w_1 \leftarrow tmp_1 \cdot tmp_2$ \\ 292513. $tmp_1 \leftarrow 2 \cdot a_2$, $tmp_1 \leftarrow a_1 + tmp_1$, $tmp_1 \leftarrow 2 \cdot tmp_1$, $tmp_1 \leftarrow tmp_1 + a_0$ \\ 292614. $tmp_2 \leftarrow 2 \cdot b_2$, $tmp_2 \leftarrow b_1 + tmp_2$, $tmp_2 \leftarrow 2 \cdot tmp_2$, $tmp_2 \leftarrow tmp_2 + b_0$ \\ 292715. $w_3 \leftarrow tmp_1 \cdot tmp_2$ \\ 292816. $tmp_1 \leftarrow a_0 + a_1$, $tmp_1 \leftarrow tmp_1 + a_2$, $tmp_2 \leftarrow b_0 + b_1$, $tmp_2 \leftarrow tmp_2 + b_2$ \\ 292917. $w_2 \leftarrow tmp_1 \cdot tmp_2$ \\ 2930\\ 2931Continued on the next page.\\ 2932\hline 2933\end{tabular} 2934\end{center} 2935\end{small} 2936\caption{Algorithm mp\_toom\_mul} 2937\end{figure} 2938 2939\newpage\begin{figure}[!here] 2940\begin{small} 2941\begin{center} 2942\begin{tabular}{l} 2943\hline Algorithm \textbf{mp\_toom\_mul} (continued). \\ 2944\textbf{Input}. mp\_int $a$ and mp\_int $b$ \\ 2945\textbf{Output}. $c \leftarrow a \cdot b $ \\ 2946\hline \\ 2947Now solve the system of equations. \\ 294818. $w_1 \leftarrow w_4 - w_1$, $w_3 \leftarrow w_3 - w_0$ \\ 294919. $w_1 \leftarrow \lfloor w_1 / 2 \rfloor$, $w_3 \leftarrow \lfloor w_3 / 2 \rfloor$ \\ 295020. $w_2 \leftarrow w_2 - w_0$, $w_2 \leftarrow w_2 - w_4$ \\ 295121. $w_1 \leftarrow w_1 - w_2$, $w_3 \leftarrow w_3 - w_2$ \\ 295222. $tmp_1 \leftarrow 8 \cdot w_0$, $w_1 \leftarrow w_1 - tmp_1$, $tmp_1 \leftarrow 8 \cdot w_4$, $w_3 \leftarrow w_3 - tmp_1$ \\ 295323. $w_2 \leftarrow 3 \cdot w_2$, $w_2 \leftarrow w_2 - w_1$, $w_2 \leftarrow w_2 - w_3$ \\ 295424. $w_1 \leftarrow w_1 - w_2$, $w_3 \leftarrow w_3 - w_2$ \\ 295525. $w_1 \leftarrow \lfloor w_1 / 3 \rfloor, w_3 \leftarrow \lfloor w_3 / 3 \rfloor$ \\ 2956\\ 2957Now substitute $\beta^k$ for $x$ by shifting $w_0, w_1, ..., w_4$. \\ 295826. for $n$ from $1$ to $4$ do \\ 2959\hspace{3mm}26.1 $w_n \leftarrow w_n \cdot \beta^{nk}$ \\ 296027. $c \leftarrow w_0 + w_1$, $c \leftarrow c + w_2$, $c \leftarrow c + w_3$, $c \leftarrow c + w_4$ \\ 296128. Return(\textit{MP\_OKAY}) \\ 2962\hline 2963\end{tabular} 2964\end{center} 2965\end{small} 2966\caption{Algorithm mp\_toom\_mul (continued)} 2967\end{figure} 2968 2969\textbf{Algorithm mp\_toom\_mul.} 2970This algorithm computes the product of two mp\_int variables $a$ and $b$ using the Toom-Cook approach. Compared to the Karatsuba multiplication, this 2971algorithm has a lower asymptotic running time of approximately $O(n^{1.464})$ but at an obvious cost in overhead. In this 2972description, several statements have been compounded to save space. The intention is that the statements are executed from left to right across 2973any given step. 2974 2975The two inputs $a$ and $b$ are first split into three $k$-digit integers $a_0, a_1, a_2$ and $b_0, b_1, b_2$ respectively. From these smaller 2976integers the coefficients of the polynomial basis representations $f(x)$ and $g(x)$ are known and can be used to find the relations required. 2977 2978The first two relations $w_0$ and $w_4$ are the points $\zeta_{0}$ and $\zeta_{\infty}$ respectively. The relation $w_1, w_2$ and $w_3$ correspond 2979to the points $16 \cdot \zeta_{1 \over 2}, \zeta_{2}$ and $\zeta_{1}$ respectively. These are found using logical shifts to independently find 2980$f(y)$ and $g(y)$ which significantly speeds up the algorithm. 2981 2982After the five relations $w_0, w_1, \ldots, w_4$ have been computed, the system they represent must be solved in order for the unknown coefficients 2983$w_1, w_2$ and $w_3$ to be isolated. The steps 18 through 25 perform the system reduction required as previously described. Each step of 2984the reduction represents the comparable matrix operation that would be performed had this been performed by pencil. For example, step 18 indicates 2985that row $1$ must be subtracted from row $4$ and simultaneously row $0$ subtracted from row $3$. 2986 2987Once the coeffients have been isolated, the polynomial $W(x) = \sum_{i=0}^{2n} w_i x^i$ is known. By substituting $\beta^{k}$ for $x$, the integer 2988result $a \cdot b$ is produced. 2989 2990EXAM,bn_mp_toom_mul.c 2991 2992The first obvious thing to note is that this algorithm is complicated. The complexity is worth it if you are multiplying very 2993large numbers. For example, a 10,000 digit multiplication takes approximaly 99,282,205 fewer single precision multiplications with 2994Toom--Cook than a Comba or baseline approach (this is a savings of more than 99$\%$). For most ``crypto'' sized numbers this 2995algorithm is not practical as Karatsuba has a much lower cutoff point. 2996 2997First we split $a$ and $b$ into three roughly equal portions. This has been accomplished (lines @40,mod@ to @69,rshd@) with 2998combinations of mp\_rshd() and mp\_mod\_2d() function calls. At this point $a = a2 \cdot \beta^2 + a1 \cdot \beta + a0$ and similiarly 2999for $b$. 3000 3001Next we compute the five points $w0, w1, w2, w3$ and $w4$. Recall that $w0$ and $w4$ can be computed directly from the portions so 3002we get those out of the way first (lines @72,mul@ and @77,mul@). Next we compute $w1, w2$ and $w3$ using Horners method. 3003 3004After this point we solve for the actual values of $w1, w2$ and $w3$ by reducing the $5 \times 5$ system which is relatively 3005straight forward. 3006 3007\subsection{Signed Multiplication} 3008Now that algorithms to handle multiplications of every useful dimensions have been developed, a rather simple finishing touch is required. So far all 3009of the multiplication algorithms have been unsigned multiplications which leaves only a signed multiplication algorithm to be established. 3010 3011\begin{figure}[!here] 3012\begin{small} 3013\begin{center} 3014\begin{tabular}{l} 3015\hline Algorithm \textbf{mp\_mul}. \\ 3016\textbf{Input}. mp\_int $a$ and mp\_int $b$ \\ 3017\textbf{Output}. $c \leftarrow a \cdot b$ \\ 3018\hline \\ 30191. If $a.sign = b.sign$ then \\ 3020\hspace{3mm}1.1 $sign = MP\_ZPOS$ \\ 30212. else \\ 3022\hspace{3mm}2.1 $sign = MP\_ZNEG$ \\ 30233. If min$(a.used, b.used) \ge TOOM\_MUL\_CUTOFF$ then \\ 3024\hspace{3mm}3.1 $c \leftarrow a \cdot b$ using algorithm mp\_toom\_mul \\ 30254. else if min$(a.used, b.used) \ge KARATSUBA\_MUL\_CUTOFF$ then \\ 3026\hspace{3mm}4.1 $c \leftarrow a \cdot b$ using algorithm mp\_karatsuba\_mul \\ 30275. else \\ 3028\hspace{3mm}5.1 $digs \leftarrow a.used + b.used + 1$ \\ 3029\hspace{3mm}5.2 If $digs < MP\_ARRAY$ and min$(a.used, b.used) \le \delta$ then \\ 3030\hspace{6mm}5.2.1 $c \leftarrow a \cdot b \mbox{ (mod }\beta^{digs}\mbox{)}$ using algorithm fast\_s\_mp\_mul\_digs. \\ 3031\hspace{3mm}5.3 else \\ 3032\hspace{6mm}5.3.1 $c \leftarrow a \cdot b \mbox{ (mod }\beta^{digs}\mbox{)}$ using algorithm s\_mp\_mul\_digs. \\ 30336. $c.sign \leftarrow sign$ \\ 30347. Return the result of the unsigned multiplication performed. \\ 3035\hline 3036\end{tabular} 3037\end{center} 3038\end{small} 3039\caption{Algorithm mp\_mul} 3040\end{figure} 3041 3042\textbf{Algorithm mp\_mul.} 3043This algorithm performs the signed multiplication of two inputs. It will make use of any of the three unsigned multiplication algorithms 3044available when the input is of appropriate size. The \textbf{sign} of the result is not set until the end of the algorithm since algorithm 3045s\_mp\_mul\_digs will clear it. 3046 3047EXAM,bn_mp_mul.c 3048 3049The implementation is rather simplistic and is not particularly noteworthy. Line @22,?@ computes the sign of the result using the ``?'' 3050operator from the C programming language. Line @37,<<@ computes $\delta$ using the fact that $1 << k$ is equal to $2^k$. 3051 3052\section{Squaring} 3053\label{sec:basesquare} 3054 3055Squaring is a special case of multiplication where both multiplicands are equal. At first it may seem like there is no significant optimization 3056available but in fact there is. Consider the multiplication of $576$ against $241$. In total there will be nine single precision multiplications 3057performed which are $1\cdot 6$, $1 \cdot 7$, $1 \cdot 5$, $4 \cdot 6$, $4 \cdot 7$, $4 \cdot 5$, $2 \cdot 6$, $2 \cdot 7$ and $2 \cdot 5$. Now consider 3058the multiplication of $123$ against $123$. The nine products are $3 \cdot 3$, $3 \cdot 2$, $3 \cdot 1$, $2 \cdot 3$, $2 \cdot 2$, $2 \cdot 1$, 3059$1 \cdot 3$, $1 \cdot 2$ and $1 \cdot 1$. On closer inspection some of the products are equivalent. For example, $3 \cdot 2 = 2 \cdot 3$ 3060and $3 \cdot 1 = 1 \cdot 3$. 3061 3062For any $n$-digit input, there are ${{\left (n^2 + n \right)}\over 2}$ possible unique single precision multiplications required compared to the $n^2$ 3063required for multiplication. The following diagram gives an example of the operations required. 3064 3065\begin{figure}[here] 3066\begin{center} 3067\begin{tabular}{ccccc|c} 3068&&1&2&3&\\ 3069$\times$ &&1&2&3&\\ 3070\hline && $3 \cdot 1$ & $3 \cdot 2$ & $3 \cdot 3$ & Row 0\\ 3071 & $2 \cdot 1$ & $2 \cdot 2$ & $2 \cdot 3$ && Row 1 \\ 3072 $1 \cdot 1$ & $1 \cdot 2$ & $1 \cdot 3$ &&& Row 2 \\ 3073\end{tabular} 3074\end{center} 3075\caption{Squaring Optimization Diagram} 3076\end{figure} 3077 3078MARK,SQUARE 3079Starting from zero and numbering the columns from right to left a very simple pattern becomes obvious. For the purposes of this discussion let $x$ 3080represent the number being squared. The first observation is that in row $k$ the $2k$'th column of the product has a $\left (x_k \right)^2$ term in it. 3081 3082The second observation is that every column $j$ in row $k$ where $j \ne 2k$ is part of a double product. Every non-square term of a column will 3083appear twice hence the name ``double product''. Every odd column is made up entirely of double products. In fact every column is made up of double 3084products and at most one square (\textit{see the exercise section}). 3085 3086The third and final observation is that for row $k$ the first unique non-square term, that is, one that hasn't already appeared in an earlier row, 3087occurs at column $2k + 1$. For example, on row $1$ of the previous squaring, column one is part of the double product with column one from row zero. 3088Column two of row one is a square and column three is the first unique column. 3089 3090\subsection{The Baseline Squaring Algorithm} 3091The baseline squaring algorithm is meant to be a catch-all squaring algorithm. It will handle any of the input sizes that the faster routines 3092will not handle. 3093 3094\begin{figure}[!here] 3095\begin{small} 3096\begin{center} 3097\begin{tabular}{l} 3098\hline Algorithm \textbf{s\_mp\_sqr}. \\ 3099\textbf{Input}. mp\_int $a$ \\ 3100\textbf{Output}. $b \leftarrow a^2$ \\ 3101\hline \\ 31021. Init a temporary mp\_int of at least $2 \cdot a.used +1$ digits. (\textit{mp\_init\_size}) \\ 31032. If step 1 failed return(\textit{MP\_MEM}) \\ 31043. $t.used \leftarrow 2 \cdot a.used + 1$ \\ 31054. For $ix$ from 0 to $a.used - 1$ do \\ 3106\hspace{3mm}Calculate the square. \\ 3107\hspace{3mm}4.1 $\hat r \leftarrow t_{2ix} + \left (a_{ix} \right )^2$ \\ 3108\hspace{3mm}4.2 $t_{2ix} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\ 3109\hspace{3mm}Calculate the double products after the square. \\ 3110\hspace{3mm}4.3 $u \leftarrow \lfloor \hat r / \beta \rfloor$ \\ 3111\hspace{3mm}4.4 For $iy$ from $ix + 1$ to $a.used - 1$ do \\ 3112\hspace{6mm}4.4.1 $\hat r \leftarrow 2 \cdot a_{ix}a_{iy} + t_{ix + iy} + u$ \\ 3113\hspace{6mm}4.4.2 $t_{ix + iy} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\ 3114\hspace{6mm}4.4.3 $u \leftarrow \lfloor \hat r / \beta \rfloor$ \\ 3115\hspace{3mm}Set the last carry. \\ 3116\hspace{3mm}4.5 While $u > 0$ do \\ 3117\hspace{6mm}4.5.1 $iy \leftarrow iy + 1$ \\ 3118\hspace{6mm}4.5.2 $\hat r \leftarrow t_{ix + iy} + u$ \\ 3119\hspace{6mm}4.5.3 $t_{ix + iy} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\ 3120\hspace{6mm}4.5.4 $u \leftarrow \lfloor \hat r / \beta \rfloor$ \\ 31215. Clamp excess digits of $t$. (\textit{mp\_clamp}) \\ 31226. Exchange $b$ and $t$. \\ 31237. Clear $t$ (\textit{mp\_clear}) \\ 31248. Return(\textit{MP\_OKAY}) \\ 3125\hline 3126\end{tabular} 3127\end{center} 3128\end{small} 3129\caption{Algorithm s\_mp\_sqr} 3130\end{figure} 3131 3132\textbf{Algorithm s\_mp\_sqr.} 3133This algorithm computes the square of an input using the three observations on squaring. It is based fairly faithfully on algorithm 14.16 of HAC 3134\cite[pp.596-597]{HAC}. Similar to algorithm s\_mp\_mul\_digs, a temporary mp\_int is allocated to hold the result of the squaring. This allows the 3135destination mp\_int to be the same as the source mp\_int. 3136 3137The outer loop of this algorithm begins on step 4. It is best to think of the outer loop as walking down the rows of the partial results, while 3138the inner loop computes the columns of the partial result. Step 4.1 and 4.2 compute the square term for each row, and step 4.3 and 4.4 propagate 3139the carry and compute the double products. 3140 3141The requirement that a mp\_word be able to represent the range $0 \le x < 2 \beta^2$ arises from this 3142very algorithm. The product $a_{ix}a_{iy}$ will lie in the range $0 \le x \le \beta^2 - 2\beta + 1$ which is obviously less than $\beta^2$ meaning that 3143when it is multiplied by two, it can be properly represented by a mp\_word. 3144 3145Similar to algorithm s\_mp\_mul\_digs, after every pass of the inner loop, the destination is correctly set to the sum of all of the partial 3146results calculated so far. This involves expensive carry propagation which will be eliminated in the next algorithm. 3147 3148EXAM,bn_s_mp_sqr.c 3149 3150Inside the outer loop (line @32,for@) the square term is calculated on line @35,r =@. The carry (line @42,>>@) has been 3151extracted from the mp\_word accumulator using a right shift. Aliases for $a_{ix}$ and $t_{ix+iy}$ are initialized 3152(lines @45,tmpx@ and @48,tmpt@) to simplify the inner loop. The doubling is performed using two 3153additions (line @57,r + r@) since it is usually faster than shifting, if not at least as fast. 3154 3155The important observation is that the inner loop does not begin at $iy = 0$ like for multiplication. As such the inner loops 3156get progressively shorter as the algorithm proceeds. This is what leads to the savings compared to using a multiplication to 3157square a number. 3158 3159\subsection{Faster Squaring by the ``Comba'' Method} 3160A major drawback to the baseline method is the requirement for single precision shifting inside the $O(n^2)$ nested loop. Squaring has an additional 3161drawback that it must double the product inside the inner loop as well. As for multiplication, the Comba technique can be used to eliminate these 3162performance hazards. 3163 3164The first obvious solution is to make an array of mp\_words which will hold all of the columns. This will indeed eliminate all of the carry 3165propagation operations from the inner loop. However, the inner product must still be doubled $O(n^2)$ times. The solution stems from the simple fact 3166that $2a + 2b + 2c = 2(a + b + c)$. That is the sum of all of the double products is equal to double the sum of all the products. For example, 3167$ab + ba + ac + ca = 2ab + 2ac = 2(ab + ac)$. 3168 3169However, we cannot simply double all of the columns, since the squares appear only once per row. The most practical solution is to have two 3170mp\_word arrays. One array will hold the squares and the other array will hold the double products. With both arrays the doubling and 3171carry propagation can be moved to a $O(n)$ work level outside the $O(n^2)$ level. In this case, we have an even simpler solution in mind. 3172 3173\newpage\begin{figure}[!here] 3174\begin{small} 3175\begin{center} 3176\begin{tabular}{l} 3177\hline Algorithm \textbf{fast\_s\_mp\_sqr}. \\ 3178\textbf{Input}. mp\_int $a$ \\ 3179\textbf{Output}. $b \leftarrow a^2$ \\ 3180\hline \\ 3181Place an array of \textbf{MP\_WARRAY} mp\_digits named $W$ on the stack. \\ 31821. If $b.alloc < 2a.used + 1$ then grow $b$ to $2a.used + 1$ digits. (\textit{mp\_grow}). \\ 31832. If step 1 failed return(\textit{MP\_MEM}). \\ 3184\\ 31853. $pa \leftarrow 2 \cdot a.used$ \\ 31864. $\hat W1 \leftarrow 0$ \\ 31875. for $ix$ from $0$ to $pa - 1$ do \\ 3188\hspace{3mm}5.1 $\_ \hat W \leftarrow 0$ \\ 3189\hspace{3mm}5.2 $ty \leftarrow \mbox{MIN}(a.used - 1, ix)$ \\ 3190\hspace{3mm}5.3 $tx \leftarrow ix - ty$ \\ 3191\hspace{3mm}5.4 $iy \leftarrow \mbox{MIN}(a.used - tx, ty + 1)$ \\ 3192\hspace{3mm}5.5 $iy \leftarrow \mbox{MIN}(iy, \lfloor \left (ty - tx + 1 \right )/2 \rfloor)$ \\ 3193\hspace{3mm}5.6 for $iz$ from $0$ to $iz - 1$ do \\ 3194\hspace{6mm}5.6.1 $\_ \hat W \leftarrow \_ \hat W + a_{tx + iz}a_{ty - iz}$ \\ 3195\hspace{3mm}5.7 $\_ \hat W \leftarrow 2 \cdot \_ \hat W + \hat W1$ \\ 3196\hspace{3mm}5.8 if $ix$ is even then \\ 3197\hspace{6mm}5.8.1 $\_ \hat W \leftarrow \_ \hat W + \left ( a_{\lfloor ix/2 \rfloor}\right )^2$ \\ 3198\hspace{3mm}5.9 $W_{ix} \leftarrow \_ \hat W (\mbox{mod }\beta)$ \\ 3199\hspace{3mm}5.10 $\hat W1 \leftarrow \lfloor \_ \hat W / \beta \rfloor$ \\ 3200\\ 32016. $oldused \leftarrow b.used$ \\ 32027. $b.used \leftarrow 2 \cdot a.used$ \\ 32038. for $ix$ from $0$ to $pa - 1$ do \\ 3204\hspace{3mm}8.1 $b_{ix} \leftarrow W_{ix}$ \\ 32059. for $ix$ from $pa$ to $oldused - 1$ do \\ 3206\hspace{3mm}9.1 $b_{ix} \leftarrow 0$ \\ 320710. Clamp excess digits from $b$. (\textit{mp\_clamp}) \\ 320811. Return(\textit{MP\_OKAY}). \\ 3209\hline 3210\end{tabular} 3211\end{center} 3212\end{small} 3213\caption{Algorithm fast\_s\_mp\_sqr} 3214\end{figure} 3215 3216\textbf{Algorithm fast\_s\_mp\_sqr.} 3217This algorithm computes the square of an input using the Comba technique. It is designed to be a replacement for algorithm 3218s\_mp\_sqr when the number of input digits is less than \textbf{MP\_WARRAY} and less than $\delta \over 2$. 3219This algorithm is very similar to the Comba multiplier except with a few key differences we shall make note of. 3220 3221First, we have an accumulator and carry variables $\_ \hat W$ and $\hat W1$ respectively. This is because the inner loop 3222products are to be doubled. If we had added the previous carry in we would be doubling too much. Next we perform an 3223addition MIN condition on $iy$ (step 5.5) to prevent overlapping digits. For example, $a_3 \cdot a_5$ is equal 3224$a_5 \cdot a_3$. Whereas in the multiplication case we would have $5 < a.used$ and $3 \ge 0$ is maintained since we double the sum 3225of the products just outside the inner loop we have to avoid doing this. This is also a good thing since we perform 3226fewer multiplications and the routine ends up being faster. 3227 3228Finally the last difference is the addition of the ``square'' term outside the inner loop (step 5.8). We add in the square 3229only to even outputs and it is the square of the term at the $\lfloor ix / 2 \rfloor$ position. 3230 3231EXAM,bn_fast_s_mp_sqr.c 3232 3233This implementation is essentially a copy of Comba multiplication with the appropriate changes added to make it faster for 3234the special case of squaring. 3235 3236\subsection{Polynomial Basis Squaring} 3237The same algorithm that performs optimal polynomial basis multiplication can be used to perform polynomial basis squaring. The minor exception 3238is that $\zeta_y = f(y)g(y)$ is actually equivalent to $\zeta_y = f(y)^2$ since $f(y) = g(y)$. Instead of performing $2n + 1$ 3239multiplications to find the $\zeta$ relations, squaring operations are performed instead. 3240 3241\subsection{Karatsuba Squaring} 3242Let $f(x) = ax + b$ represent the polynomial basis representation of a number to square. 3243Let $h(x) = \left ( f(x) \right )^2$ represent the square of the polynomial. The Karatsuba equation can be modified to square a 3244number with the following equation. 3245 3246\begin{equation} 3247h(x) = a^2x^2 + \left ((a + b)^2 - (a^2 + b^2) \right )x + b^2 3248\end{equation} 3249 3250Upon closer inspection this equation only requires the calculation of three half-sized squares: $a^2$, $b^2$ and $(a + b)^2$. As in 3251Karatsuba multiplication, this algorithm can be applied recursively on the input and will achieve an asymptotic running time of 3252$O \left ( n^{lg(3)} \right )$. 3253 3254If the asymptotic times of Karatsuba squaring and multiplication are the same, why not simply use the multiplication algorithm 3255instead? The answer to this arises from the cutoff point for squaring. As in multiplication there exists a cutoff point, at which the 3256time required for a Comba based squaring and a Karatsuba based squaring meet. Due to the overhead inherent in the Karatsuba method, the cutoff 3257point is fairly high. For example, on an AMD Athlon XP processor with $\beta = 2^{28}$, the cutoff point is around 127 digits. 3258 3259Consider squaring a 200 digit number with this technique. It will be split into two 100 digit halves which are subsequently squared. 3260The 100 digit halves will not be squared using Karatsuba, but instead using the faster Comba based squaring algorithm. If Karatsuba multiplication 3261were used instead, the 100 digit numbers would be squared with a slower Comba based multiplication. 3262 3263\newpage\begin{figure}[!here] 3264\begin{small} 3265\begin{center} 3266\begin{tabular}{l} 3267\hline Algorithm \textbf{mp\_karatsuba\_sqr}. \\ 3268\textbf{Input}. mp\_int $a$ \\ 3269\textbf{Output}. $b \leftarrow a^2$ \\ 3270\hline \\ 32711. Initialize the following temporary mp\_ints: $x0$, $x1$, $t1$, $t2$, $x0x0$ and $x1x1$. \\ 32722. If any of the initializations on step 1 failed return(\textit{MP\_MEM}). \\ 3273\\ 3274Split the input. e.g. $a = x1\beta^B + x0$ \\ 32753. $B \leftarrow \lfloor a.used / 2 \rfloor$ \\ 32764. $x0 \leftarrow a \mbox{ (mod }\beta^B\mbox{)}$ (\textit{mp\_mod\_2d}) \\ 32775. $x1 \leftarrow \lfloor a / \beta^B \rfloor$ (\textit{mp\_lshd}) \\ 3278\\ 3279Calculate the three squares. \\ 32806. $x0x0 \leftarrow x0^2$ (\textit{mp\_sqr}) \\ 32817. $x1x1 \leftarrow x1^2$ \\ 32828. $t1 \leftarrow x1 + x0$ (\textit{s\_mp\_add}) \\ 32839. $t1 \leftarrow t1^2$ \\ 3284\\ 3285Compute the middle term. \\ 328610. $t2 \leftarrow x0x0 + x1x1$ (\textit{s\_mp\_add}) \\ 328711. $t1 \leftarrow t1 - t2$ \\ 3288\\ 3289Compute final product. \\ 329012. $t1 \leftarrow t1\beta^B$ (\textit{mp\_lshd}) \\ 329113. $x1x1 \leftarrow x1x1\beta^{2B}$ \\ 329214. $t1 \leftarrow t1 + x0x0$ \\ 329315. $b \leftarrow t1 + x1x1$ \\ 329416. Return(\textit{MP\_OKAY}). \\ 3295\hline 3296\end{tabular} 3297\end{center} 3298\end{small} 3299\caption{Algorithm mp\_karatsuba\_sqr} 3300\end{figure} 3301 3302\textbf{Algorithm mp\_karatsuba\_sqr.} 3303This algorithm computes the square of an input $a$ using the Karatsuba technique. This algorithm is very similar to the Karatsuba based 3304multiplication algorithm with the exception that the three half-size multiplications have been replaced with three half-size squarings. 3305 3306The radix point for squaring is simply placed exactly in the middle of the digits when the input has an odd number of digits, otherwise it is 3307placed just below the middle. Step 3, 4 and 5 compute the two halves required using $B$ 3308as the radix point. The first two squares in steps 6 and 7 are rather straightforward while the last square is of a more compact form. 3309 3310By expanding $\left (x1 + x0 \right )^2$, the $x1^2$ and $x0^2$ terms in the middle disappear, that is $(x0 - x1)^2 - (x1^2 + x0^2) = 2 \cdot x0 \cdot x1$. 3311Now if $5n$ single precision additions and a squaring of $n$-digits is faster than multiplying two $n$-digit numbers and doubling then 3312this method is faster. Assuming no further recursions occur, the difference can be estimated with the following inequality. 3313 3314Let $p$ represent the cost of a single precision addition and $q$ the cost of a single precision multiplication both in terms of time\footnote{Or 3315machine clock cycles.}. 3316 3317\begin{equation} 33185pn +{{q(n^2 + n)} \over 2} \le pn + qn^2 3319\end{equation} 3320 3321For example, on an AMD Athlon XP processor $p = {1 \over 3}$ and $q = 6$. This implies that the following inequality should hold. 3322\begin{center} 3323\begin{tabular}{rcl} 3324${5n \over 3} + 3n^2 + 3n$ & $<$ & ${n \over 3} + 6n^2$ \\ 3325${5 \over 3} + 3n + 3$ & $<$ & ${1 \over 3} + 6n$ \\ 3326${13 \over 9}$ & $<$ & $n$ \\ 3327\end{tabular} 3328\end{center} 3329 3330This results in a cutoff point around $n = 2$. As a consequence it is actually faster to compute the middle term the ``long way'' on processors 3331where multiplication is substantially slower\footnote{On the Athlon there is a 1:17 ratio between clock cycles for addition and multiplication. On 3332the Intel P4 processor this ratio is 1:29 making this method even more beneficial. The only common exception is the ARMv4 processor which has a 3333ratio of 1:7. } than simpler operations such as addition. 3334 3335EXAM,bn_mp_karatsuba_sqr.c 3336 3337This implementation is largely based on the implementation of algorithm mp\_karatsuba\_mul. It uses the same inline style to copy and 3338shift the input into the two halves. The loop from line @54,{@ to line @70,}@ has been modified since only one input exists. The \textbf{used} 3339count of both $x0$ and $x1$ is fixed up and $x0$ is clamped before the calculations begin. At this point $x1$ and $x0$ are valid equivalents 3340to the respective halves as if mp\_rshd and mp\_mod\_2d had been used. 3341 3342By inlining the copy and shift operations the cutoff point for Karatsuba multiplication can be lowered. On the Athlon the cutoff point 3343is exactly at the point where Comba squaring can no longer be used (\textit{128 digits}). On slower processors such as the Intel P4 3344it is actually below the Comba limit (\textit{at 110 digits}). 3345 3346This routine uses the same error trap coding style as mp\_karatsuba\_sqr. As the temporary variables are initialized errors are 3347redirected to the error trap higher up. If the algorithm completes without error the error code is set to \textbf{MP\_OKAY} and 3348mp\_clears are executed normally. 3349 3350\subsection{Toom-Cook Squaring} 3351The Toom-Cook squaring algorithm mp\_toom\_sqr is heavily based on the algorithm mp\_toom\_mul with the exception that squarings are used 3352instead of multiplication to find the five relations. The reader is encouraged to read the description of the latter algorithm and try to 3353derive their own Toom-Cook squaring algorithm. 3354 3355\subsection{High Level Squaring} 3356\newpage\begin{figure}[!here] 3357\begin{small} 3358\begin{center} 3359\begin{tabular}{l} 3360\hline Algorithm \textbf{mp\_sqr}. \\ 3361\textbf{Input}. mp\_int $a$ \\ 3362\textbf{Output}. $b \leftarrow a^2$ \\ 3363\hline \\ 33641. If $a.used \ge TOOM\_SQR\_CUTOFF$ then \\ 3365\hspace{3mm}1.1 $b \leftarrow a^2$ using algorithm mp\_toom\_sqr \\ 33662. else if $a.used \ge KARATSUBA\_SQR\_CUTOFF$ then \\ 3367\hspace{3mm}2.1 $b \leftarrow a^2$ using algorithm mp\_karatsuba\_sqr \\ 33683. else \\ 3369\hspace{3mm}3.1 $digs \leftarrow a.used + b.used + 1$ \\ 3370\hspace{3mm}3.2 If $digs < MP\_ARRAY$ and $a.used \le \delta$ then \\ 3371\hspace{6mm}3.2.1 $b \leftarrow a^2$ using algorithm fast\_s\_mp\_sqr. \\ 3372\hspace{3mm}3.3 else \\ 3373\hspace{6mm}3.3.1 $b \leftarrow a^2$ using algorithm s\_mp\_sqr. \\ 33744. $b.sign \leftarrow MP\_ZPOS$ \\ 33755. Return the result of the unsigned squaring performed. \\ 3376\hline 3377\end{tabular} 3378\end{center} 3379\end{small} 3380\caption{Algorithm mp\_sqr} 3381\end{figure} 3382 3383\textbf{Algorithm mp\_sqr.} 3384This algorithm computes the square of the input using one of four different algorithms. If the input is very large and has at least 3385\textbf{TOOM\_SQR\_CUTOFF} or \textbf{KARATSUBA\_SQR\_CUTOFF} digits then either the Toom-Cook or the Karatsuba Squaring algorithm is used. If 3386neither of the polynomial basis algorithms should be used then either the Comba or baseline algorithm is used. 3387 3388EXAM,bn_mp_sqr.c 3389 3390\section*{Exercises} 3391\begin{tabular}{cl} 3392$\left [ 3 \right ] $ & Devise an efficient algorithm for selection of the radix point to handle inputs \\ 3393 & that have different number of digits in Karatsuba multiplication. \\ 3394 & \\ 3395$\left [ 2 \right ] $ & In ~SQUARE~ the fact that every column of a squaring is made up \\ 3396 & of double products and at most one square is stated. Prove this statement. \\ 3397 & \\ 3398$\left [ 3 \right ] $ & Prove the equation for Karatsuba squaring. \\ 3399 & \\ 3400$\left [ 1 \right ] $ & Prove that Karatsuba squaring requires $O \left (n^{lg(3)} \right )$ time. \\ 3401 & \\ 3402$\left [ 2 \right ] $ & Determine the minimal ratio between addition and multiplication clock cycles \\ 3403 & required for equation $6.7$ to be true. \\ 3404 & \\ 3405$\left [ 3 \right ] $ & Implement a threaded version of Comba multiplication (and squaring) where you \\ 3406 & compute subsets of the columns in each thread. Determine a cutoff point where \\ 3407 & it is effective and add the logic to mp\_mul() and mp\_sqr(). \\ 3408 &\\ 3409$\left [ 4 \right ] $ & Same as the previous but also modify the Karatsuba and Toom-Cook. You must \\ 3410 & increase the throughput of mp\_exptmod() for random odd moduli in the range \\ 3411 & $512 \ldots 4096$ bits significantly ($> 2x$) to complete this challenge. \\ 3412 & \\ 3413\end{tabular} 3414 3415\chapter{Modular Reduction} 3416MARK,REDUCTION 3417\section{Basics of Modular Reduction} 3418\index{modular residue} 3419Modular reduction is an operation that arises quite often within public key cryptography algorithms and various number theoretic algorithms, 3420such as factoring. Modular reduction algorithms are the third class of algorithms of the ``multipliers'' set. A number $a$ is said to be \textit{reduced} 3421modulo another number $b$ by finding the remainder of the division $a/b$. Full integer division with remainder is a topic to be covered 3422in~\ref{sec:division}. 3423 3424Modular reduction is equivalent to solving for $r$ in the following equation. $a = bq + r$ where $q = \lfloor a/b \rfloor$. The result 3425$r$ is said to be ``congruent to $a$ modulo $b$'' which is also written as $r \equiv a \mbox{ (mod }b\mbox{)}$. In other vernacular $r$ is known as the 3426``modular residue'' which leads to ``quadratic residue''\footnote{That's fancy talk for $b \equiv a^2 \mbox{ (mod }p\mbox{)}$.} and 3427other forms of residues. 3428 3429Modular reductions are normally used to create either finite groups, rings or fields. The most common usage for performance driven modular reductions 3430is in modular exponentiation algorithms. That is to compute $d = a^b \mbox{ (mod }c\mbox{)}$ as fast as possible. This operation is used in the 3431RSA and Diffie-Hellman public key algorithms, for example. Modular multiplication and squaring also appears as a fundamental operation in 3432elliptic curve cryptographic algorithms. As will be discussed in the subsequent chapter there exist fast algorithms for computing modular 3433exponentiations without having to perform (\textit{in this example}) $b - 1$ multiplications. These algorithms will produce partial results in the 3434range $0 \le x < c^2$ which can be taken advantage of to create several efficient algorithms. They have also been used to create redundancy check 3435algorithms known as CRCs, error correction codes such as Reed-Solomon and solve a variety of number theoeretic problems. 3436 3437\section{The Barrett Reduction} 3438The Barrett reduction algorithm \cite{BARRETT} was inspired by fast division algorithms which multiply by the reciprocal to emulate 3439division. Barretts observation was that the residue $c$ of $a$ modulo $b$ is equal to 3440 3441\begin{equation} 3442c = a - b \cdot \lfloor a/b \rfloor 3443\end{equation} 3444 3445Since algorithms such as modular exponentiation would be using the same modulus extensively, typical DSP\footnote{It is worth noting that Barrett's paper 3446targeted the DSP56K processor.} intuition would indicate the next step would be to replace $a/b$ by a multiplication by the reciprocal. However, 3447DSP intuition on its own will not work as these numbers are considerably larger than the precision of common DSP floating point data types. 3448It would take another common optimization to optimize the algorithm. 3449 3450\subsection{Fixed Point Arithmetic} 3451The trick used to optimize the above equation is based on a technique of emulating floating point data types with fixed precision integers. Fixed 3452point arithmetic would become very popular as it greatly optimize the ``3d-shooter'' genre of games in the mid 1990s when floating point units were 3453fairly slow if not unavailable. The idea behind fixed point arithmetic is to take a normal $k$-bit integer data type and break it into $p$-bit 3454integer and a $q$-bit fraction part (\textit{where $p+q = k$}). 3455 3456In this system a $k$-bit integer $n$ would actually represent $n/2^q$. For example, with $q = 4$ the integer $n = 37$ would actually represent the 3457value $2.3125$. To multiply two fixed point numbers the integers are multiplied using traditional arithmetic and subsequently normalized by 3458moving the implied decimal point back to where it should be. For example, with $q = 4$ to multiply the integers $9$ and $5$ they must be converted 3459to fixed point first by multiplying by $2^q$. Let $a = 9(2^q)$ represent the fixed point representation of $9$ and $b = 5(2^q)$ represent the 3460fixed point representation of $5$. The product $ab$ is equal to $45(2^{2q})$ which when normalized by dividing by $2^q$ produces $45(2^q)$. 3461 3462This technique became popular since a normal integer multiplication and logical shift right are the only required operations to perform a multiplication 3463of two fixed point numbers. Using fixed point arithmetic, division can be easily approximated by multiplying by the reciprocal. If $2^q$ is 3464equivalent to one than $2^q/b$ is equivalent to the fixed point approximation of $1/b$ using real arithmetic. Using this fact dividing an integer 3465$a$ by another integer $b$ can be achieved with the following expression. 3466 3467\begin{equation} 3468\lfloor a / b \rfloor \mbox{ }\approx\mbox{ } \lfloor (a \cdot \lfloor 2^q / b \rfloor)/2^q \rfloor 3469\end{equation} 3470 3471The precision of the division is proportional to the value of $q$. If the divisor $b$ is used frequently as is the case with 3472modular exponentiation pre-computing $2^q/b$ will allow a division to be performed with a multiplication and a right shift. Both operations 3473are considerably faster than division on most processors. 3474 3475Consider dividing $19$ by $5$. The correct result is $\lfloor 19/5 \rfloor = 3$. With $q = 3$ the reciprocal is $\lfloor 2^q/5 \rfloor = 1$ which 3476leads to a product of $19$ which when divided by $2^q$ produces $2$. However, with $q = 4$ the reciprocal is $\lfloor 2^q/5 \rfloor = 3$ and 3477the result of the emulated division is $\lfloor 3 \cdot 19 / 2^q \rfloor = 3$ which is correct. The value of $2^q$ must be close to or ideally 3478larger than the dividend. In effect if $a$ is the dividend then $q$ should allow $0 \le \lfloor a/2^q \rfloor \le 1$ in order for this approach 3479to work correctly. Plugging this form of divison into the original equation the following modular residue equation arises. 3480 3481\begin{equation} 3482c = a - b \cdot \lfloor (a \cdot \lfloor 2^q / b \rfloor)/2^q \rfloor 3483\end{equation} 3484 3485Using the notation from \cite{BARRETT} the value of $\lfloor 2^q / b \rfloor$ will be represented by the $\mu$ symbol. Using the $\mu$ 3486variable also helps re-inforce the idea that it is meant to be computed once and re-used. 3487 3488\begin{equation} 3489c = a - b \cdot \lfloor (a \cdot \mu)/2^q \rfloor 3490\end{equation} 3491 3492Provided that $2^q \ge a$ this algorithm will produce a quotient that is either exactly correct or off by a value of one. In the context of Barrett 3493reduction the value of $a$ is bound by $0 \le a \le (b - 1)^2$ meaning that $2^q \ge b^2$ is sufficient to ensure the reciprocal will have enough 3494precision. 3495 3496Let $n$ represent the number of digits in $b$. This algorithm requires approximately $2n^2$ single precision multiplications to produce the quotient and 3497another $n^2$ single precision multiplications to find the residue. In total $3n^2$ single precision multiplications are required to 3498reduce the number. 3499 3500For example, if $b = 1179677$ and $q = 41$ ($2^q > b^2$), then the reciprocal $\mu$ is equal to $\lfloor 2^q / b \rfloor = 1864089$. Consider reducing 3501$a = 180388626447$ modulo $b$ using the above reduction equation. The quotient using the new formula is $\lfloor (a \cdot \mu) / 2^q \rfloor = 152913$. 3502By subtracting $152913b$ from $a$ the correct residue $a \equiv 677346 \mbox{ (mod }b\mbox{)}$ is found. 3503 3504\subsection{Choosing a Radix Point} 3505Using the fixed point representation a modular reduction can be performed with $3n^2$ single precision multiplications. If that were the best 3506that could be achieved a full division\footnote{A division requires approximately $O(2cn^2)$ single precision multiplications for a small value of $c$. 3507See~\ref{sec:division} for further details.} might as well be used in its place. The key to optimizing the reduction is to reduce the precision of 3508the initial multiplication that finds the quotient. 3509 3510Let $a$ represent the number of which the residue is sought. Let $b$ represent the modulus used to find the residue. Let $m$ represent 3511the number of digits in $b$. For the purposes of this discussion we will assume that the number of digits in $a$ is $2m$, which is generally true if 3512two $m$-digit numbers have been multiplied. Dividing $a$ by $b$ is the same as dividing a $2m$ digit integer by a $m$ digit integer. Digits below the 3513$m - 1$'th digit of $a$ will contribute at most a value of $1$ to the quotient because $\beta^k < b$ for any $0 \le k \le m - 1$. Another way to 3514express this is by re-writing $a$ as two parts. If $a' \equiv a \mbox{ (mod }b^m\mbox{)}$ and $a'' = a - a'$ then 3515${a \over b} \equiv {{a' + a''} \over b}$ which is equivalent to ${a' \over b} + {a'' \over b}$. Since $a'$ is bound to be less than $b$ the quotient 3516is bound by $0 \le {a' \over b} < 1$. 3517 3518Since the digits of $a'$ do not contribute much to the quotient the observation is that they might as well be zero. However, if the digits 3519``might as well be zero'' they might as well not be there in the first place. Let $q_0 = \lfloor a/\beta^{m-1} \rfloor$ represent the input 3520with the irrelevant digits trimmed. Now the modular reduction is trimmed to the almost equivalent equation 3521 3522\begin{equation} 3523c = a - b \cdot \lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor 3524\end{equation} 3525 3526Note that the original divisor $2^q$ has been replaced with $\beta^{m+1}$ where in this case $q$ is a multiple of $lg(\beta)$. Also note that the 3527exponent on the divisor when added to the amount $q_0$ was shifted by equals $2m$. If the optimization had not been performed the divisor 3528would have the exponent $2m$ so in the end the exponents do ``add up''. Using the above equation the quotient 3529$\lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor$ can be off from the true quotient by at most two. The original fixed point quotient can be off 3530by as much as one (\textit{provided the radix point is chosen suitably}) and now that the lower irrelevent digits have been trimmed the quotient 3531can be off by an additional value of one for a total of at most two. This implies that 3532$0 \le a - b \cdot \lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor < 3b$. By first subtracting $b$ times the quotient and then conditionally subtracting 3533$b$ once or twice the residue is found. 3534 3535The quotient is now found using $(m + 1)(m) = m^2 + m$ single precision multiplications and the residue with an additional $m^2$ single 3536precision multiplications, ignoring the subtractions required. In total $2m^2 + m$ single precision multiplications are required to find the residue. 3537This is considerably faster than the original attempt. 3538 3539For example, let $\beta = 10$ represent the radix of the digits. Let $b = 9999$ represent the modulus which implies $m = 4$. Let $a = 99929878$ 3540represent the value of which the residue is desired. In this case $q = 8$ since $10^7 < 9999^2$ meaning that $\mu = \lfloor \beta^{q}/b \rfloor = 10001$. 3541With the new observation the multiplicand for the quotient is equal to $q_0 = \lfloor a / \beta^{m - 1} \rfloor = 99929$. The quotient is then 3542$\lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor = 9993$. Subtracting $9993b$ from $a$ and the correct residue $a \equiv 9871 \mbox{ (mod }b\mbox{)}$ 3543is found. 3544 3545\subsection{Trimming the Quotient} 3546So far the reduction algorithm has been optimized from $3m^2$ single precision multiplications down to $2m^2 + m$ single precision multiplications. As 3547it stands now the algorithm is already fairly fast compared to a full integer division algorithm. However, there is still room for 3548optimization. 3549 3550After the first multiplication inside the quotient ($q_0 \cdot \mu$) the value is shifted right by $m + 1$ places effectively nullifying the lower 3551half of the product. It would be nice to be able to remove those digits from the product to effectively cut down the number of single precision 3552multiplications. If the number of digits in the modulus $m$ is far less than $\beta$ a full product is not required for the algorithm to work properly. 3553In fact the lower $m - 2$ digits will not affect the upper half of the product at all and do not need to be computed. 3554 3555The value of $\mu$ is a $m$-digit number and $q_0$ is a $m + 1$ digit number. Using a full multiplier $(m + 1)(m) = m^2 + m$ single precision 3556multiplications would be required. Using a multiplier that will only produce digits at and above the $m - 1$'th digit reduces the number 3557of single precision multiplications to ${m^2 + m} \over 2$ single precision multiplications. 3558 3559\subsection{Trimming the Residue} 3560After the quotient has been calculated it is used to reduce the input. As previously noted the algorithm is not exact and it can be off by a small 3561multiple of the modulus, that is $0 \le a - b \cdot \lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor < 3b$. If $b$ is $m$ digits than the 3562result of reduction equation is a value of at most $m + 1$ digits (\textit{provided $3 < \beta$}) implying that the upper $m - 1$ digits are 3563implicitly zero. 3564 3565The next optimization arises from this very fact. Instead of computing $b \cdot \lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor$ using a full 3566$O(m^2)$ multiplication algorithm only the lower $m+1$ digits of the product have to be computed. Similarly the value of $a$ can 3567be reduced modulo $\beta^{m+1}$ before the multiple of $b$ is subtracted which simplifes the subtraction as well. A multiplication that produces 3568only the lower $m+1$ digits requires ${m^2 + 3m - 2} \over 2$ single precision multiplications. 3569 3570With both optimizations in place the algorithm is the algorithm Barrett proposed. It requires $m^2 + 2m - 1$ single precision multiplications which 3571is considerably faster than the straightforward $3m^2$ method. 3572 3573\subsection{The Barrett Algorithm} 3574\newpage\begin{figure}[!here] 3575\begin{small} 3576\begin{center} 3577\begin{tabular}{l} 3578\hline Algorithm \textbf{mp\_reduce}. \\ 3579\textbf{Input}. mp\_int $a$, mp\_int $b$ and $\mu = \lfloor \beta^{2m}/b \rfloor, m = \lceil lg_{\beta}(b) \rceil, (0 \le a < b^2, b > 1)$ \\ 3580\textbf{Output}. $a \mbox{ (mod }b\mbox{)}$ \\ 3581\hline \\ 3582Let $m$ represent the number of digits in $b$. \\ 35831. Make a copy of $a$ and store it in $q$. (\textit{mp\_init\_copy}) \\ 35842. $q \leftarrow \lfloor q / \beta^{m - 1} \rfloor$ (\textit{mp\_rshd}) \\ 3585\\ 3586Produce the quotient. \\ 35873. $q \leftarrow q \cdot \mu$ (\textit{note: only produce digits at or above $m-1$}) \\ 35884. $q \leftarrow \lfloor q / \beta^{m + 1} \rfloor$ \\ 3589\\ 3590Subtract the multiple of modulus from the input. \\ 35915. $a \leftarrow a \mbox{ (mod }\beta^{m+1}\mbox{)}$ (\textit{mp\_mod\_2d}) \\ 35926. $q \leftarrow q \cdot b \mbox{ (mod }\beta^{m+1}\mbox{)}$ (\textit{s\_mp\_mul\_digs}) \\ 35937. $a \leftarrow a - q$ (\textit{mp\_sub}) \\ 3594\\ 3595Add $\beta^{m+1}$ if a carry occured. \\ 35968. If $a < 0$ then (\textit{mp\_cmp\_d}) \\ 3597\hspace{3mm}8.1 $q \leftarrow 1$ (\textit{mp\_set}) \\ 3598\hspace{3mm}8.2 $q \leftarrow q \cdot \beta^{m+1}$ (\textit{mp\_lshd}) \\ 3599\hspace{3mm}8.3 $a \leftarrow a + q$ \\ 3600\\ 3601Now subtract the modulus if the residue is too large (e.g. quotient too small). \\ 36029. While $a \ge b$ do (\textit{mp\_cmp}) \\ 3603\hspace{3mm}9.1 $c \leftarrow a - b$ \\ 360410. Clear $q$. \\ 360511. Return(\textit{MP\_OKAY}) \\ 3606\hline 3607\end{tabular} 3608\end{center} 3609\end{small} 3610\caption{Algorithm mp\_reduce} 3611\end{figure} 3612 3613\textbf{Algorithm mp\_reduce.} 3614This algorithm will reduce the input $a$ modulo $b$ in place using the Barrett algorithm. It is loosely based on algorithm 14.42 of HAC 3615\cite[pp. 602]{HAC} which is based on the paper from Paul Barrett \cite{BARRETT}. The algorithm has several restrictions and assumptions which must 3616be adhered to for the algorithm to work. 3617 3618First the modulus $b$ is assumed to be positive and greater than one. If the modulus were less than or equal to one than subtracting 3619a multiple of it would either accomplish nothing or actually enlarge the input. The input $a$ must be in the range $0 \le a < b^2$ in order 3620for the quotient to have enough precision. If $a$ is the product of two numbers that were already reduced modulo $b$, this will not be a problem. 3621Technically the algorithm will still work if $a \ge b^2$ but it will take much longer to finish. The value of $\mu$ is passed as an argument to this 3622algorithm and is assumed to be calculated and stored before the algorithm is used. 3623 3624Recall that the multiplication for the quotient on step 3 must only produce digits at or above the $m-1$'th position. An algorithm called 3625$s\_mp\_mul\_high\_digs$ which has not been presented is used to accomplish this task. The algorithm is based on $s\_mp\_mul\_digs$ except that 3626instead of stopping at a given level of precision it starts at a given level of precision. This optimal algorithm can only be used if the number 3627of digits in $b$ is very much smaller than $\beta$. 3628 3629While it is known that 3630$a \ge b \cdot \lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor$ only the lower $m+1$ digits are being used to compute the residue, so an implied 3631``borrow'' from the higher digits might leave a negative result. After the multiple of the modulus has been subtracted from $a$ the residue must be 3632fixed up in case it is negative. The invariant $\beta^{m+1}$ must be added to the residue to make it positive again. 3633 3634The while loop at step 9 will subtract $b$ until the residue is less than $b$. If the algorithm is performed correctly this step is 3635performed at most twice, and on average once. However, if $a \ge b^2$ than it will iterate substantially more times than it should. 3636 3637EXAM,bn_mp_reduce.c 3638 3639The first multiplication that determines the quotient can be performed by only producing the digits from $m - 1$ and up. This essentially halves 3640the number of single precision multiplications required. However, the optimization is only safe if $\beta$ is much larger than the number of digits 3641in the modulus. In the source code this is evaluated on lines @36,if@ to @44,}@ where algorithm s\_mp\_mul\_high\_digs is used when it is 3642safe to do so. 3643 3644\subsection{The Barrett Setup Algorithm} 3645In order to use algorithm mp\_reduce the value of $\mu$ must be calculated in advance. Ideally this value should be computed once and stored for 3646future use so that the Barrett algorithm can be used without delay. 3647 3648\newpage\begin{figure}[!here] 3649\begin{small} 3650\begin{center} 3651\begin{tabular}{l} 3652\hline Algorithm \textbf{mp\_reduce\_setup}. \\ 3653\textbf{Input}. mp\_int $a$ ($a > 1$) \\ 3654\textbf{Output}. $\mu \leftarrow \lfloor \beta^{2m}/a \rfloor$ \\ 3655\hline \\ 36561. $\mu \leftarrow 2^{2 \cdot lg(\beta) \cdot m}$ (\textit{mp\_2expt}) \\ 36572. $\mu \leftarrow \lfloor \mu / b \rfloor$ (\textit{mp\_div}) \\ 36583. Return(\textit{MP\_OKAY}) \\ 3659\hline 3660\end{tabular} 3661\end{center} 3662\end{small} 3663\caption{Algorithm mp\_reduce\_setup} 3664\end{figure} 3665 3666\textbf{Algorithm mp\_reduce\_setup.} 3667This algorithm computes the reciprocal $\mu$ required for Barrett reduction. First $\beta^{2m}$ is calculated as $2^{2 \cdot lg(\beta) \cdot m}$ which 3668is equivalent and much faster. The final value is computed by taking the integer quotient of $\lfloor \mu / b \rfloor$. 3669 3670EXAM,bn_mp_reduce_setup.c 3671 3672This simple routine calculates the reciprocal $\mu$ required by Barrett reduction. Note the extended usage of algorithm mp\_div where the variable 3673which would received the remainder is passed as NULL. As will be discussed in~\ref{sec:division} the division routine allows both the quotient and the 3674remainder to be passed as NULL meaning to ignore the value. 3675 3676\section{The Montgomery Reduction} 3677Montgomery reduction\footnote{Thanks to Niels Ferguson for his insightful explanation of the algorithm.} \cite{MONT} is by far the most interesting 3678form of reduction in common use. It computes a modular residue which is not actually equal to the residue of the input yet instead equal to a 3679residue times a constant. However, as perplexing as this may sound the algorithm is relatively simple and very efficient. 3680 3681Throughout this entire section the variable $n$ will represent the modulus used to form the residue. As will be discussed shortly the value of 3682$n$ must be odd. The variable $x$ will represent the quantity of which the residue is sought. Similar to the Barrett algorithm the input 3683is restricted to $0 \le x < n^2$. To begin the description some simple number theory facts must be established. 3684 3685\textbf{Fact 1.} Adding $n$ to $x$ does not change the residue since in effect it adds one to the quotient $\lfloor x / n \rfloor$. Another way 3686to explain this is that $n$ is (\textit{or multiples of $n$ are}) congruent to zero modulo $n$. Adding zero will not change the value of the residue. 3687 3688\textbf{Fact 2.} If $x$ is even then performing a division by two in $\Z$ is congruent to $x \cdot 2^{-1} \mbox{ (mod }n\mbox{)}$. Actually 3689this is an application of the fact that if $x$ is evenly divisible by any $k \in \Z$ then division in $\Z$ will be congruent to 3690multiplication by $k^{-1}$ modulo $n$. 3691 3692From these two simple facts the following simple algorithm can be derived. 3693 3694\newpage\begin{figure}[!here] 3695\begin{small} 3696\begin{center} 3697\begin{tabular}{l} 3698\hline Algorithm \textbf{Montgomery Reduction}. \\ 3699\textbf{Input}. Integer $x$, $n$ and $k$ \\ 3700\textbf{Output}. $2^{-k}x \mbox{ (mod }n\mbox{)}$ \\ 3701\hline \\ 37021. for $t$ from $1$ to $k$ do \\ 3703\hspace{3mm}1.1 If $x$ is odd then \\ 3704\hspace{6mm}1.1.1 $x \leftarrow x + n$ \\ 3705\hspace{3mm}1.2 $x \leftarrow x/2$ \\ 37062. Return $x$. \\ 3707\hline 3708\end{tabular} 3709\end{center} 3710\end{small} 3711\caption{Algorithm Montgomery Reduction} 3712\end{figure} 3713 3714The algorithm reduces the input one bit at a time using the two congruencies stated previously. Inside the loop $n$, which is odd, is 3715added to $x$ if $x$ is odd. This forces $x$ to be even which allows the division by two in $\Z$ to be congruent to a modular division by two. Since 3716$x$ is assumed to be initially much larger than $n$ the addition of $n$ will contribute an insignificant magnitude to $x$. Let $r$ represent the 3717final result of the Montgomery algorithm. If $k > lg(n)$ and $0 \le x < n^2$ then the final result is limited to 3718$0 \le r < \lfloor x/2^k \rfloor + n$. As a result at most a single subtraction is required to get the residue desired. 3719 3720\begin{figure}[here] 3721\begin{small} 3722\begin{center} 3723\begin{tabular}{|c|l|} 3724\hline \textbf{Step number ($t$)} & \textbf{Result ($x$)} \\ 3725\hline $1$ & $x + n = 5812$, $x/2 = 2906$ \\ 3726\hline $2$ & $x/2 = 1453$ \\ 3727\hline $3$ & $x + n = 1710$, $x/2 = 855$ \\ 3728\hline $4$ & $x + n = 1112$, $x/2 = 556$ \\ 3729\hline $5$ & $x/2 = 278$ \\ 3730\hline $6$ & $x/2 = 139$ \\ 3731\hline $7$ & $x + n = 396$, $x/2 = 198$ \\ 3732\hline $8$ & $x/2 = 99$ \\ 3733\hline $9$ & $x + n = 356$, $x/2 = 178$ \\ 3734\hline 3735\end{tabular} 3736\end{center} 3737\end{small} 3738\caption{Example of Montgomery Reduction (I)} 3739\label{fig:MONT1} 3740\end{figure} 3741 3742Consider the example in figure~\ref{fig:MONT1} which reduces $x = 5555$ modulo $n = 257$ when $k = 9$ (note $\beta^k = 512$ which is larger than $n$). The result of 3743the algorithm $r = 178$ is congruent to the value of $2^{-9} \cdot 5555 \mbox{ (mod }257\mbox{)}$. When $r$ is multiplied by $2^9$ modulo $257$ the correct residue 3744$r \equiv 158$ is produced. 3745 3746Let $k = \lfloor lg(n) \rfloor + 1$ represent the number of bits in $n$. The current algorithm requires $2k^2$ single precision shifts 3747and $k^2$ single precision additions. At this rate the algorithm is most certainly slower than Barrett reduction and not terribly useful. 3748Fortunately there exists an alternative representation of the algorithm. 3749 3750\begin{figure}[!here] 3751\begin{small} 3752\begin{center} 3753\begin{tabular}{l} 3754\hline Algorithm \textbf{Montgomery Reduction} (modified I). \\ 3755\textbf{Input}. Integer $x$, $n$ and $k$ ($2^k > n$) \\ 3756\textbf{Output}. $2^{-k}x \mbox{ (mod }n\mbox{)}$ \\ 3757\hline \\ 37581. for $t$ from $1$ to $k$ do \\ 3759\hspace{3mm}1.1 If the $t$'th bit of $x$ is one then \\ 3760\hspace{6mm}1.1.1 $x \leftarrow x + 2^tn$ \\ 37612. Return $x/2^k$. \\ 3762\hline 3763\end{tabular} 3764\end{center} 3765\end{small} 3766\caption{Algorithm Montgomery Reduction (modified I)} 3767\end{figure} 3768 3769This algorithm is equivalent since $2^tn$ is a multiple of $n$ and the lower $k$ bits of $x$ are zero by step 2. The number of single 3770precision shifts has now been reduced from $2k^2$ to $k^2 + k$ which is only a small improvement. 3771 3772\begin{figure}[here] 3773\begin{small} 3774\begin{center} 3775\begin{tabular}{|c|l|r|} 3776\hline \textbf{Step number ($t$)} & \textbf{Result ($x$)} & \textbf{Result ($x$) in Binary} \\ 3777\hline -- & $5555$ & $1010110110011$ \\ 3778\hline $1$ & $x + 2^{0}n = 5812$ & $1011010110100$ \\ 3779\hline $2$ & $5812$ & $1011010110100$ \\ 3780\hline $3$ & $x + 2^{2}n = 6840$ & $1101010111000$ \\ 3781\hline $4$ & $x + 2^{3}n = 8896$ & $10001011000000$ \\ 3782\hline $5$ & $8896$ & $10001011000000$ \\ 3783\hline $6$ & $8896$ & $10001011000000$ \\ 3784\hline $7$ & $x + 2^{6}n = 25344$ & $110001100000000$ \\ 3785\hline $8$ & $25344$ & $110001100000000$ \\ 3786\hline $9$ & $x + 2^{7}n = 91136$ & $10110010000000000$ \\ 3787\hline -- & $x/2^k = 178$ & \\ 3788\hline 3789\end{tabular} 3790\end{center} 3791\end{small} 3792\caption{Example of Montgomery Reduction (II)} 3793\label{fig:MONT2} 3794\end{figure} 3795 3796Figure~\ref{fig:MONT2} demonstrates the modified algorithm reducing $x = 5555$ modulo $n = 257$ with $k = 9$. 3797With this algorithm a single shift right at the end is the only right shift required to reduce the input instead of $k$ right shifts inside the 3798loop. Note that for the iterations $t = 2, 5, 6$ and $8$ where the result $x$ is not changed. In those iterations the $t$'th bit of $x$ is 3799zero and the appropriate multiple of $n$ does not need to be added to force the $t$'th bit of the result to zero. 3800 3801\subsection{Digit Based Montgomery Reduction} 3802Instead of computing the reduction on a bit-by-bit basis it is actually much faster to compute it on digit-by-digit basis. Consider the 3803previous algorithm re-written to compute the Montgomery reduction in this new fashion. 3804 3805\begin{figure}[!here] 3806\begin{small} 3807\begin{center} 3808\begin{tabular}{l} 3809\hline Algorithm \textbf{Montgomery Reduction} (modified II). \\ 3810\textbf{Input}. Integer $x$, $n$ and $k$ ($\beta^k > n$) \\ 3811\textbf{Output}. $\beta^{-k}x \mbox{ (mod }n\mbox{)}$ \\ 3812\hline \\ 38131. for $t$ from $0$ to $k - 1$ do \\ 3814\hspace{3mm}1.1 $x \leftarrow x + \mu n \beta^t$ \\ 38152. Return $x/\beta^k$. \\ 3816\hline 3817\end{tabular} 3818\end{center} 3819\end{small} 3820\caption{Algorithm Montgomery Reduction (modified II)} 3821\end{figure} 3822 3823The value $\mu n \beta^t$ is a multiple of the modulus $n$ meaning that it will not change the residue. If the first digit of 3824the value $\mu n \beta^t$ equals the negative (modulo $\beta$) of the $t$'th digit of $x$ then the addition will result in a zero digit. This 3825problem breaks down to solving the following congruency. 3826 3827\begin{center} 3828\begin{tabular}{rcl} 3829$x_t + \mu n_0$ & $\equiv$ & $0 \mbox{ (mod }\beta\mbox{)}$ \\ 3830$\mu n_0$ & $\equiv$ & $-x_t \mbox{ (mod }\beta\mbox{)}$ \\ 3831$\mu$ & $\equiv$ & $-x_t/n_0 \mbox{ (mod }\beta\mbox{)}$ \\ 3832\end{tabular} 3833\end{center} 3834 3835In each iteration of the loop on step 1 a new value of $\mu$ must be calculated. The value of $-1/n_0 \mbox{ (mod }\beta\mbox{)}$ is used 3836extensively in this algorithm and should be precomputed. Let $\rho$ represent the negative of the modular inverse of $n_0$ modulo $\beta$. 3837 3838For example, let $\beta = 10$ represent the radix. Let $n = 17$ represent the modulus which implies $k = 2$ and $\rho \equiv 7$. Let $x = 33$ 3839represent the value to reduce. 3840 3841\newpage\begin{figure} 3842\begin{center} 3843\begin{tabular}{|c|c|c|} 3844\hline \textbf{Step ($t$)} & \textbf{Value of $x$} & \textbf{Value of $\mu$} \\ 3845\hline -- & $33$ & --\\ 3846\hline $0$ & $33 + \mu n = 50$ & $1$ \\ 3847\hline $1$ & $50 + \mu n \beta = 900$ & $5$ \\ 3848\hline 3849\end{tabular} 3850\end{center} 3851\caption{Example of Montgomery Reduction} 3852\end{figure} 3853 3854The final result $900$ is then divided by $\beta^k$ to produce the final result $9$. The first observation is that $9 \nequiv x \mbox{ (mod }n\mbox{)}$ 3855which implies the result is not the modular residue of $x$ modulo $n$. However, recall that the residue is actually multiplied by $\beta^{-k}$ in 3856the algorithm. To get the true residue the value must be multiplied by $\beta^k$. In this case $\beta^k \equiv 15 \mbox{ (mod }n\mbox{)}$ and 3857the correct residue is $9 \cdot 15 \equiv 16 \mbox{ (mod }n\mbox{)}$. 3858 3859\subsection{Baseline Montgomery Reduction} 3860The baseline Montgomery reduction algorithm will produce the residue for any size input. It is designed to be a catch-all algororithm for 3861Montgomery reductions. 3862 3863\newpage\begin{figure}[!here] 3864\begin{small} 3865\begin{center} 3866\begin{tabular}{l} 3867\hline Algorithm \textbf{mp\_montgomery\_reduce}. \\ 3868\textbf{Input}. mp\_int $x$, mp\_int $n$ and a digit $\rho \equiv -1/n_0 \mbox{ (mod }n\mbox{)}$. \\ 3869\hspace{11.5mm}($0 \le x < n^2, n > 1, (n, \beta) = 1, \beta^k > n$) \\ 3870\textbf{Output}. $\beta^{-k}x \mbox{ (mod }n\mbox{)}$ \\ 3871\hline \\ 38721. $digs \leftarrow 2n.used + 1$ \\ 38732. If $digs < MP\_ARRAY$ and $m.used < \delta$ then \\ 3874\hspace{3mm}2.1 Use algorithm fast\_mp\_montgomery\_reduce instead. \\ 3875\\ 3876Setup $x$ for the reduction. \\ 38773. If $x.alloc < digs$ then grow $x$ to $digs$ digits. \\ 38784. $x.used \leftarrow digs$ \\ 3879\\ 3880Eliminate the lower $k$ digits. \\ 38815. For $ix$ from $0$ to $k - 1$ do \\ 3882\hspace{3mm}5.1 $\mu \leftarrow x_{ix} \cdot \rho \mbox{ (mod }\beta\mbox{)}$ \\ 3883\hspace{3mm}5.2 $u \leftarrow 0$ \\ 3884\hspace{3mm}5.3 For $iy$ from $0$ to $k - 1$ do \\ 3885\hspace{6mm}5.3.1 $\hat r \leftarrow \mu n_{iy} + x_{ix + iy} + u$ \\ 3886\hspace{6mm}5.3.2 $x_{ix + iy} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\ 3887\hspace{6mm}5.3.3 $u \leftarrow \lfloor \hat r / \beta \rfloor$ \\ 3888\hspace{3mm}5.4 While $u > 0$ do \\ 3889\hspace{6mm}5.4.1 $iy \leftarrow iy + 1$ \\ 3890\hspace{6mm}5.4.2 $x_{ix + iy} \leftarrow x_{ix + iy} + u$ \\ 3891\hspace{6mm}5.4.3 $u \leftarrow \lfloor x_{ix+iy} / \beta \rfloor$ \\ 3892\hspace{6mm}5.4.4 $x_{ix + iy} \leftarrow x_{ix+iy} \mbox{ (mod }\beta\mbox{)}$ \\ 3893\\ 3894Divide by $\beta^k$ and fix up as required. \\ 38956. $x \leftarrow \lfloor x / \beta^k \rfloor$ \\ 38967. If $x \ge n$ then \\ 3897\hspace{3mm}7.1 $x \leftarrow x - n$ \\ 38988. Return(\textit{MP\_OKAY}). \\ 3899\hline 3900\end{tabular} 3901\end{center} 3902\end{small} 3903\caption{Algorithm mp\_montgomery\_reduce} 3904\end{figure} 3905 3906\textbf{Algorithm mp\_montgomery\_reduce.} 3907This algorithm reduces the input $x$ modulo $n$ in place using the Montgomery reduction algorithm. The algorithm is loosely based 3908on algorithm 14.32 of \cite[pp.601]{HAC} except it merges the multiplication of $\mu n \beta^t$ with the addition in the inner loop. The 3909restrictions on this algorithm are fairly easy to adapt to. First $0 \le x < n^2$ bounds the input to numbers in the same range as 3910for the Barrett algorithm. Additionally if $n > 1$ and $n$ is odd there will exist a modular inverse $\rho$. $\rho$ must be calculated in 3911advance of this algorithm. Finally the variable $k$ is fixed and a pseudonym for $n.used$. 3912 3913Step 2 decides whether a faster Montgomery algorithm can be used. It is based on the Comba technique meaning that there are limits on 3914the size of the input. This algorithm is discussed in ~COMBARED~. 3915 3916Step 5 is the main reduction loop of the algorithm. The value of $\mu$ is calculated once per iteration in the outer loop. The inner loop 3917calculates $x + \mu n \beta^{ix}$ by multiplying $\mu n$ and adding the result to $x$ shifted by $ix$ digits. Both the addition and 3918multiplication are performed in the same loop to save time and memory. Step 5.4 will handle any additional carries that escape the inner loop. 3919 3920Using a quick inspection this algorithm requires $n$ single precision multiplications for the outer loop and $n^2$ single precision multiplications 3921in the inner loop. In total $n^2 + n$ single precision multiplications which compares favourably to Barrett at $n^2 + 2n - 1$ single precision 3922multiplications. 3923 3924EXAM,bn_mp_montgomery_reduce.c 3925 3926This is the baseline implementation of the Montgomery reduction algorithm. Lines @30,digs@ to @35,}@ determine if the Comba based 3927routine can be used instead. Line @47,mu@ computes the value of $\mu$ for that particular iteration of the outer loop. 3928 3929The multiplication $\mu n \beta^{ix}$ is performed in one step in the inner loop. The alias $tmpx$ refers to the $ix$'th digit of $x$ and 3930the alias $tmpn$ refers to the modulus $n$. 3931 3932\subsection{Faster ``Comba'' Montgomery Reduction} 3933MARK,COMBARED 3934 3935The Montgomery reduction requires fewer single precision multiplications than a Barrett reduction, however it is much slower due to the serial 3936nature of the inner loop. The Barrett reduction algorithm requires two slightly modified multipliers which can be implemented with the Comba 3937technique. The Montgomery reduction algorithm cannot directly use the Comba technique to any significant advantage since the inner loop calculates 3938a $k \times 1$ product $k$ times. 3939 3940The biggest obstacle is that at the $ix$'th iteration of the outer loop the value of $x_{ix}$ is required to calculate $\mu$. This means the 3941carries from $0$ to $ix - 1$ must have been propagated upwards to form a valid $ix$'th digit. The solution as it turns out is very simple. 3942Perform a Comba like multiplier and inside the outer loop just after the inner loop fix up the $ix + 1$'th digit by forwarding the carry. 3943 3944With this change in place the Montgomery reduction algorithm can be performed with a Comba style multiplication loop which substantially increases 3945the speed of the algorithm. 3946 3947\newpage\begin{figure}[!here] 3948\begin{small} 3949\begin{center} 3950\begin{tabular}{l} 3951\hline Algorithm \textbf{fast\_mp\_montgomery\_reduce}. \\ 3952\textbf{Input}. mp\_int $x$, mp\_int $n$ and a digit $\rho \equiv -1/n_0 \mbox{ (mod }n\mbox{)}$. \\ 3953\hspace{11.5mm}($0 \le x < n^2, n > 1, (n, \beta) = 1, \beta^k > n$) \\ 3954\textbf{Output}. $\beta^{-k}x \mbox{ (mod }n\mbox{)}$ \\ 3955\hline \\ 3956Place an array of \textbf{MP\_WARRAY} mp\_word variables called $\hat W$ on the stack. \\ 39571. if $x.alloc < n.used + 1$ then grow $x$ to $n.used + 1$ digits. \\ 3958Copy the digits of $x$ into the array $\hat W$ \\ 39592. For $ix$ from $0$ to $x.used - 1$ do \\ 3960\hspace{3mm}2.1 $\hat W_{ix} \leftarrow x_{ix}$ \\ 39613. For $ix$ from $x.used$ to $2n.used - 1$ do \\ 3962\hspace{3mm}3.1 $\hat W_{ix} \leftarrow 0$ \\ 3963Elimiate the lower $k$ digits. \\ 39644. for $ix$ from $0$ to $n.used - 1$ do \\ 3965\hspace{3mm}4.1 $\mu \leftarrow \hat W_{ix} \cdot \rho \mbox{ (mod }\beta\mbox{)}$ \\ 3966\hspace{3mm}4.2 For $iy$ from $0$ to $n.used - 1$ do \\ 3967\hspace{6mm}4.2.1 $\hat W_{iy + ix} \leftarrow \hat W_{iy + ix} + \mu \cdot n_{iy}$ \\ 3968\hspace{3mm}4.3 $\hat W_{ix + 1} \leftarrow \hat W_{ix + 1} + \lfloor \hat W_{ix} / \beta \rfloor$ \\ 3969Propagate carries upwards. \\ 39705. for $ix$ from $n.used$ to $2n.used + 1$ do \\ 3971\hspace{3mm}5.1 $\hat W_{ix + 1} \leftarrow \hat W_{ix + 1} + \lfloor \hat W_{ix} / \beta \rfloor$ \\ 3972Shift right and reduce modulo $\beta$ simultaneously. \\ 39736. for $ix$ from $0$ to $n.used + 1$ do \\ 3974\hspace{3mm}6.1 $x_{ix} \leftarrow \hat W_{ix + n.used} \mbox{ (mod }\beta\mbox{)}$ \\ 3975Zero excess digits and fixup $x$. \\ 39767. if $x.used > n.used + 1$ then do \\ 3977\hspace{3mm}7.1 for $ix$ from $n.used + 1$ to $x.used - 1$ do \\ 3978\hspace{6mm}7.1.1 $x_{ix} \leftarrow 0$ \\ 39798. $x.used \leftarrow n.used + 1$ \\ 39809. Clamp excessive digits of $x$. \\ 398110. If $x \ge n$ then \\ 3982\hspace{3mm}10.1 $x \leftarrow x - n$ \\ 398311. Return(\textit{MP\_OKAY}). \\ 3984\hline 3985\end{tabular} 3986\end{center} 3987\end{small} 3988\caption{Algorithm fast\_mp\_montgomery\_reduce} 3989\end{figure} 3990 3991\textbf{Algorithm fast\_mp\_montgomery\_reduce.} 3992This algorithm will compute the Montgomery reduction of $x$ modulo $n$ using the Comba technique. It is on most computer platforms significantly 3993faster than algorithm mp\_montgomery\_reduce and algorithm mp\_reduce (\textit{Barrett reduction}). The algorithm has the same restrictions 3994on the input as the baseline reduction algorithm. An additional two restrictions are imposed on this algorithm. The number of digits $k$ in the 3995the modulus $n$ must not violate $MP\_WARRAY > 2k +1$ and $n < \delta$. When $\beta = 2^{28}$ this algorithm can be used to reduce modulo 3996a modulus of at most $3,556$ bits in length. 3997 3998As in the other Comba reduction algorithms there is a $\hat W$ array which stores the columns of the product. It is initially filled with the 3999contents of $x$ with the excess digits zeroed. The reduction loop is very similar the to the baseline loop at heart. The multiplication on step 40004.1 can be single precision only since $ab \mbox{ (mod }\beta\mbox{)} \equiv (a \mbox{ mod }\beta)(b \mbox{ mod }\beta)$. Some multipliers such 4001as those on the ARM processors take a variable length time to complete depending on the number of bytes of result it must produce. By performing 4002a single precision multiplication instead half the amount of time is spent. 4003 4004Also note that digit $\hat W_{ix}$ must have the carry from the $ix - 1$'th digit propagated upwards in order for this to work. That is what step 40054.3 will do. In effect over the $n.used$ iterations of the outer loop the $n.used$'th lower columns all have the their carries propagated forwards. Note 4006how the upper bits of those same words are not reduced modulo $\beta$. This is because those values will be discarded shortly and there is no 4007point. 4008 4009Step 5 will propagate the remainder of the carries upwards. On step 6 the columns are reduced modulo $\beta$ and shifted simultaneously as they are 4010stored in the destination $x$. 4011 4012EXAM,bn_fast_mp_montgomery_reduce.c 4013 4014The $\hat W$ array is first filled with digits of $x$ on line @49,for@ then the rest of the digits are zeroed on line @54,for@. Both loops share 4015the same alias variables to make the code easier to read. 4016 4017The value of $\mu$ is calculated in an interesting fashion. First the value $\hat W_{ix}$ is reduced modulo $\beta$ and cast to a mp\_digit. This 4018forces the compiler to use a single precision multiplication and prevents any concerns about loss of precision. Line @101,>>@ fixes the carry 4019for the next iteration of the loop by propagating the carry from $\hat W_{ix}$ to $\hat W_{ix+1}$. 4020 4021The for loop on line @113,for@ propagates the rest of the carries upwards through the columns. The for loop on line @126,for@ reduces the columns 4022modulo $\beta$ and shifts them $k$ places at the same time. The alias $\_ \hat W$ actually refers to the array $\hat W$ starting at the $n.used$'th 4023digit, that is $\_ \hat W_{t} = \hat W_{n.used + t}$. 4024 4025\subsection{Montgomery Setup} 4026To calculate the variable $\rho$ a relatively simple algorithm will be required. 4027 4028\begin{figure}[!here] 4029\begin{small} 4030\begin{center} 4031\begin{tabular}{l} 4032\hline Algorithm \textbf{mp\_montgomery\_setup}. \\ 4033\textbf{Input}. mp\_int $n$ ($n > 1$ and $(n, 2) = 1$) \\ 4034\textbf{Output}. $\rho \equiv -1/n_0 \mbox{ (mod }\beta\mbox{)}$ \\ 4035\hline \\ 40361. $b \leftarrow n_0$ \\ 40372. If $b$ is even return(\textit{MP\_VAL}) \\ 40383. $x \leftarrow (((b + 2) \mbox{ AND } 4) << 1) + b$ \\ 40394. for $k$ from 0 to $\lceil lg(lg(\beta)) \rceil - 2$ do \\ 4040\hspace{3mm}4.1 $x \leftarrow x \cdot (2 - bx)$ \\ 40415. $\rho \leftarrow \beta - x \mbox{ (mod }\beta\mbox{)}$ \\ 40426. Return(\textit{MP\_OKAY}). \\ 4043\hline 4044\end{tabular} 4045\end{center} 4046\end{small} 4047\caption{Algorithm mp\_montgomery\_setup} 4048\end{figure} 4049 4050\textbf{Algorithm mp\_montgomery\_setup.} 4051This algorithm will calculate the value of $\rho$ required within the Montgomery reduction algorithms. It uses a very interesting trick 4052to calculate $1/n_0$ when $\beta$ is a power of two. 4053 4054EXAM,bn_mp_montgomery_setup.c 4055 4056This source code computes the value of $\rho$ required to perform Montgomery reduction. It has been modified to avoid performing excess 4057multiplications when $\beta$ is not the default 28-bits. 4058 4059\section{The Diminished Radix Algorithm} 4060The Diminished Radix method of modular reduction \cite{DRMET} is a fairly clever technique which can be more efficient than either the Barrett 4061or Montgomery methods for certain forms of moduli. The technique is based on the following simple congruence. 4062 4063\begin{equation} 4064(x \mbox{ mod } n) + k \lfloor x / n \rfloor \equiv x \mbox{ (mod }(n - k)\mbox{)} 4065\end{equation} 4066 4067This observation was used in the MMB \cite{MMB} block cipher to create a diffusion primitive. It used the fact that if $n = 2^{31}$ and $k=1$ that 4068then a x86 multiplier could produce the 62-bit product and use the ``shrd'' instruction to perform a double-precision right shift. The proof 4069of the above equation is very simple. First write $x$ in the product form. 4070 4071\begin{equation} 4072x = qn + r 4073\end{equation} 4074 4075Now reduce both sides modulo $(n - k)$. 4076 4077\begin{equation} 4078x \equiv qk + r \mbox{ (mod }(n-k)\mbox{)} 4079\end{equation} 4080 4081The variable $n$ reduces modulo $n - k$ to $k$. By putting $q = \lfloor x/n \rfloor$ and $r = x \mbox{ mod } n$ 4082into the equation the original congruence is reproduced, thus concluding the proof. The following algorithm is based on this observation. 4083 4084\begin{figure}[!here] 4085\begin{small} 4086\begin{center} 4087\begin{tabular}{l} 4088\hline Algorithm \textbf{Diminished Radix Reduction}. \\ 4089\textbf{Input}. Integer $x$, $n$, $k$ \\ 4090\textbf{Output}. $x \mbox{ mod } (n - k)$ \\ 4091\hline \\ 40921. $q \leftarrow \lfloor x / n \rfloor$ \\ 40932. $q \leftarrow k \cdot q$ \\ 40943. $x \leftarrow x \mbox{ (mod }n\mbox{)}$ \\ 40954. $x \leftarrow x + q$ \\ 40965. If $x \ge (n - k)$ then \\ 4097\hspace{3mm}5.1 $x \leftarrow x - (n - k)$ \\ 4098\hspace{3mm}5.2 Goto step 1. \\ 40996. Return $x$ \\ 4100\hline 4101\end{tabular} 4102\end{center} 4103\end{small} 4104\caption{Algorithm Diminished Radix Reduction} 4105\label{fig:DR} 4106\end{figure} 4107 4108This algorithm will reduce $x$ modulo $n - k$ and return the residue. If $0 \le x < (n - k)^2$ then the algorithm will loop almost always 4109once or twice and occasionally three times. For simplicity sake the value of $x$ is bounded by the following simple polynomial. 4110 4111\begin{equation} 41120 \le x < n^2 + k^2 - 2nk 4113\end{equation} 4114 4115The true bound is $0 \le x < (n - k - 1)^2$ but this has quite a few more terms. The value of $q$ after step 1 is bounded by the following. 4116 4117\begin{equation} 4118q < n - 2k - k^2/n 4119\end{equation} 4120 4121Since $k^2$ is going to be considerably smaller than $n$ that term will always be zero. The value of $x$ after step 3 is bounded trivially as 4122$0 \le x < n$. By step four the sum $x + q$ is bounded by 4123 4124\begin{equation} 41250 \le q + x < (k + 1)n - 2k^2 - 1 4126\end{equation} 4127 4128With a second pass $q$ will be loosely bounded by $0 \le q < k^2$ after step 2 while $x$ will still be loosely bounded by $0 \le x < n$ after step 3. After the second pass it is highly unlike that the 4129sum in step 4 will exceed $n - k$. In practice fewer than three passes of the algorithm are required to reduce virtually every input in the 4130range $0 \le x < (n - k - 1)^2$. 4131 4132\begin{figure} 4133\begin{small} 4134\begin{center} 4135\begin{tabular}{|l|} 4136\hline 4137$x = 123456789, n = 256, k = 3$ \\ 4138\hline $q \leftarrow \lfloor x/n \rfloor = 482253$ \\ 4139$q \leftarrow q*k = 1446759$ \\ 4140$x \leftarrow x \mbox{ mod } n = 21$ \\ 4141$x \leftarrow x + q = 1446780$ \\ 4142$x \leftarrow x - (n - k) = 1446527$ \\ 4143\hline 4144$q \leftarrow \lfloor x/n \rfloor = 5650$ \\ 4145$q \leftarrow q*k = 16950$ \\ 4146$x \leftarrow x \mbox{ mod } n = 127$ \\ 4147$x \leftarrow x + q = 17077$ \\ 4148$x \leftarrow x - (n - k) = 16824$ \\ 4149\hline 4150$q \leftarrow \lfloor x/n \rfloor = 65$ \\ 4151$q \leftarrow q*k = 195$ \\ 4152$x \leftarrow x \mbox{ mod } n = 184$ \\ 4153$x \leftarrow x + q = 379$ \\ 4154$x \leftarrow x - (n - k) = 126$ \\ 4155\hline 4156\end{tabular} 4157\end{center} 4158\end{small} 4159\caption{Example Diminished Radix Reduction} 4160\label{fig:EXDR} 4161\end{figure} 4162 4163Figure~\ref{fig:EXDR} demonstrates the reduction of $x = 123456789$ modulo $n - k = 253$ when $n = 256$ and $k = 3$. Note that even while $x$ 4164is considerably larger than $(n - k - 1)^2 = 63504$ the algorithm still converges on the modular residue exceedingly fast. In this case only 4165three passes were required to find the residue $x \equiv 126$. 4166 4167 4168\subsection{Choice of Moduli} 4169On the surface this algorithm looks like a very expensive algorithm. It requires a couple of subtractions followed by multiplication and other 4170modular reductions. The usefulness of this algorithm becomes exceedingly clear when an appropriate modulus is chosen. 4171 4172Division in general is a very expensive operation to perform. The one exception is when the division is by a power of the radix of representation used. 4173Division by ten for example is simple for pencil and paper mathematics since it amounts to shifting the decimal place to the right. Similarly division 4174by two (\textit{or powers of two}) is very simple for binary computers to perform. It would therefore seem logical to choose $n$ of the form $2^p$ 4175which would imply that $\lfloor x / n \rfloor$ is a simple shift of $x$ right $p$ bits. 4176 4177However, there is one operation related to division of power of twos that is even faster than this. If $n = \beta^p$ then the division may be 4178performed by moving whole digits to the right $p$ places. In practice division by $\beta^p$ is much faster than division by $2^p$ for any $p$. 4179Also with the choice of $n = \beta^p$ reducing $x$ modulo $n$ merely requires zeroing the digits above the $p-1$'th digit of $x$. 4180 4181Throughout the next section the term ``restricted modulus'' will refer to a modulus of the form $\beta^p - k$ whereas the term ``unrestricted 4182modulus'' will refer to a modulus of the form $2^p - k$. The word ``restricted'' in this case refers to the fact that it is based on the 4183$2^p$ logic except $p$ must be a multiple of $lg(\beta)$. 4184 4185\subsection{Choice of $k$} 4186Now that division and reduction (\textit{step 1 and 3 of figure~\ref{fig:DR}}) have been optimized to simple digit operations the multiplication by $k$ 4187in step 2 is the most expensive operation. Fortunately the choice of $k$ is not terribly limited. For all intents and purposes it might 4188as well be a single digit. The smaller the value of $k$ is the faster the algorithm will be. 4189 4190\subsection{Restricted Diminished Radix Reduction} 4191The restricted Diminished Radix algorithm can quickly reduce an input modulo a modulus of the form $n = \beta^p - k$. This algorithm can reduce 4192an input $x$ within the range $0 \le x < n^2$ using only a couple passes of the algorithm demonstrated in figure~\ref{fig:DR}. The implementation 4193of this algorithm has been optimized to avoid additional overhead associated with a division by $\beta^p$, the multiplication by $k$ or the addition 4194of $x$ and $q$. The resulting algorithm is very efficient and can lead to substantial improvements over Barrett and Montgomery reduction when modular 4195exponentiations are performed. 4196 4197\newpage\begin{figure}[!here] 4198\begin{small} 4199\begin{center} 4200\begin{tabular}{l} 4201\hline Algorithm \textbf{mp\_dr\_reduce}. \\ 4202\textbf{Input}. mp\_int $x$, $n$ and a mp\_digit $k = \beta - n_0$ \\ 4203\hspace{11.5mm}($0 \le x < n^2$, $n > 1$, $0 < k < \beta$) \\ 4204\textbf{Output}. $x \mbox{ mod } n$ \\ 4205\hline \\ 42061. $m \leftarrow n.used$ \\ 42072. If $x.alloc < 2m$ then grow $x$ to $2m$ digits. \\ 42083. $\mu \leftarrow 0$ \\ 42094. for $i$ from $0$ to $m - 1$ do \\ 4210\hspace{3mm}4.1 $\hat r \leftarrow k \cdot x_{m+i} + x_{i} + \mu$ \\ 4211\hspace{3mm}4.2 $x_{i} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\ 4212\hspace{3mm}4.3 $\mu \leftarrow \lfloor \hat r / \beta \rfloor$ \\ 42135. $x_{m} \leftarrow \mu$ \\ 42146. for $i$ from $m + 1$ to $x.used - 1$ do \\ 4215\hspace{3mm}6.1 $x_{i} \leftarrow 0$ \\ 42167. Clamp excess digits of $x$. \\ 42178. If $x \ge n$ then \\ 4218\hspace{3mm}8.1 $x \leftarrow x - n$ \\ 4219\hspace{3mm}8.2 Goto step 3. \\ 42209. Return(\textit{MP\_OKAY}). \\ 4221\hline 4222\end{tabular} 4223\end{center} 4224\end{small} 4225\caption{Algorithm mp\_dr\_reduce} 4226\end{figure} 4227 4228\textbf{Algorithm mp\_dr\_reduce.} 4229This algorithm will perform the Dimished Radix reduction of $x$ modulo $n$. It has similar restrictions to that of the Barrett reduction 4230with the addition that $n$ must be of the form $n = \beta^m - k$ where $0 < k <\beta$. 4231 4232This algorithm essentially implements the pseudo-code in figure~\ref{fig:DR} except with a slight optimization. The division by $\beta^m$, multiplication by $k$ 4233and addition of $x \mbox{ mod }\beta^m$ are all performed simultaneously inside the loop on step 4. The division by $\beta^m$ is emulated by accessing 4234the term at the $m+i$'th position which is subsequently multiplied by $k$ and added to the term at the $i$'th position. After the loop the $m$'th 4235digit is set to the carry and the upper digits are zeroed. Steps 5 and 6 emulate the reduction modulo $\beta^m$ that should have happend to 4236$x$ before the addition of the multiple of the upper half. 4237 4238At step 8 if $x$ is still larger than $n$ another pass of the algorithm is required. First $n$ is subtracted from $x$ and then the algorithm resumes 4239at step 3. 4240 4241EXAM,bn_mp_dr_reduce.c 4242 4243The first step is to grow $x$ as required to $2m$ digits since the reduction is performed in place on $x$. The label on line @49,top:@ is where 4244the algorithm will resume if further reduction passes are required. In theory it could be placed at the top of the function however, the size of 4245the modulus and question of whether $x$ is large enough are invariant after the first pass meaning that it would be a waste of time. 4246 4247The aliases $tmpx1$ and $tmpx2$ refer to the digits of $x$ where the latter is offset by $m$ digits. By reading digits from $x$ offset by $m$ digits 4248a division by $\beta^m$ can be simulated virtually for free. The loop on line @61,for@ performs the bulk of the work (\textit{corresponds to step 4 of algorithm 7.11}) 4249in this algorithm. 4250 4251By line @68,mu@ the pointer $tmpx1$ points to the $m$'th digit of $x$ which is where the final carry will be placed. Similarly by line @71,for@ the 4252same pointer will point to the $m+1$'th digit where the zeroes will be placed. 4253 4254Since the algorithm is only valid if both $x$ and $n$ are greater than zero an unsigned comparison suffices to determine if another pass is required. 4255With the same logic at line @82,sub@ the value of $x$ is known to be greater than or equal to $n$ meaning that an unsigned subtraction can be used 4256as well. Since the destination of the subtraction is the larger of the inputs the call to algorithm s\_mp\_sub cannot fail and the return code 4257does not need to be checked. 4258 4259\subsubsection{Setup} 4260To setup the restricted Diminished Radix algorithm the value $k = \beta - n_0$ is required. This algorithm is not really complicated but provided for 4261completeness. 4262 4263\begin{figure}[!here] 4264\begin{small} 4265\begin{center} 4266\begin{tabular}{l} 4267\hline Algorithm \textbf{mp\_dr\_setup}. \\ 4268\textbf{Input}. mp\_int $n$ \\ 4269\textbf{Output}. $k = \beta - n_0$ \\ 4270\hline \\ 42711. $k \leftarrow \beta - n_0$ \\ 4272\hline 4273\end{tabular} 4274\end{center} 4275\end{small} 4276\caption{Algorithm mp\_dr\_setup} 4277\end{figure} 4278 4279EXAM,bn_mp_dr_setup.c 4280 4281\subsubsection{Modulus Detection} 4282Another algorithm which will be useful is the ability to detect a restricted Diminished Radix modulus. An integer is said to be 4283of restricted Diminished Radix form if all of the digits are equal to $\beta - 1$ except the trailing digit which may be any value. 4284 4285\begin{figure}[!here] 4286\begin{small} 4287\begin{center} 4288\begin{tabular}{l} 4289\hline Algorithm \textbf{mp\_dr\_is\_modulus}. \\ 4290\textbf{Input}. mp\_int $n$ \\ 4291\textbf{Output}. $1$ if $n$ is in D.R form, $0$ otherwise \\ 4292\hline 42931. If $n.used < 2$ then return($0$). \\ 42942. for $ix$ from $1$ to $n.used - 1$ do \\ 4295\hspace{3mm}2.1 If $n_{ix} \ne \beta - 1$ return($0$). \\ 42963. Return($1$). \\ 4297\hline 4298\end{tabular} 4299\end{center} 4300\end{small} 4301\caption{Algorithm mp\_dr\_is\_modulus} 4302\end{figure} 4303 4304\textbf{Algorithm mp\_dr\_is\_modulus.} 4305This algorithm determines if a value is in Diminished Radix form. Step 1 rejects obvious cases where fewer than two digits are 4306in the mp\_int. Step 2 tests all but the first digit to see if they are equal to $\beta - 1$. If the algorithm manages to get to 4307step 3 then $n$ must be of Diminished Radix form. 4308 4309EXAM,bn_mp_dr_is_modulus.c 4310 4311\subsection{Unrestricted Diminished Radix Reduction} 4312The unrestricted Diminished Radix algorithm allows modular reductions to be performed when the modulus is of the form $2^p - k$. This algorithm 4313is a straightforward adaptation of algorithm~\ref{fig:DR}. 4314 4315In general the restricted Diminished Radix reduction algorithm is much faster since it has considerably lower overhead. However, this new 4316algorithm is much faster than either Montgomery or Barrett reduction when the moduli are of the appropriate form. 4317 4318\begin{figure}[!here] 4319\begin{small} 4320\begin{center} 4321\begin{tabular}{l} 4322\hline Algorithm \textbf{mp\_reduce\_2k}. \\ 4323\textbf{Input}. mp\_int $a$ and $n$. mp\_digit $k$ \\ 4324\hspace{11.5mm}($a \ge 0$, $n > 1$, $0 < k < \beta$, $n + k$ is a power of two) \\ 4325\textbf{Output}. $a \mbox{ (mod }n\mbox{)}$ \\ 4326\hline 43271. $p \leftarrow \lceil lg(n) \rceil$ (\textit{mp\_count\_bits}) \\ 43282. While $a \ge n$ do \\ 4329\hspace{3mm}2.1 $q \leftarrow \lfloor a / 2^p \rfloor$ (\textit{mp\_div\_2d}) \\ 4330\hspace{3mm}2.2 $a \leftarrow a \mbox{ (mod }2^p\mbox{)}$ (\textit{mp\_mod\_2d}) \\ 4331\hspace{3mm}2.3 $q \leftarrow q \cdot k$ (\textit{mp\_mul\_d}) \\ 4332\hspace{3mm}2.4 $a \leftarrow a - q$ (\textit{s\_mp\_sub}) \\ 4333\hspace{3mm}2.5 If $a \ge n$ then do \\ 4334\hspace{6mm}2.5.1 $a \leftarrow a - n$ \\ 43353. Return(\textit{MP\_OKAY}). \\ 4336\hline 4337\end{tabular} 4338\end{center} 4339\end{small} 4340\caption{Algorithm mp\_reduce\_2k} 4341\end{figure} 4342 4343\textbf{Algorithm mp\_reduce\_2k.} 4344This algorithm quickly reduces an input $a$ modulo an unrestricted Diminished Radix modulus $n$. Division by $2^p$ is emulated with a right 4345shift which makes the algorithm fairly inexpensive to use. 4346 4347EXAM,bn_mp_reduce_2k.c 4348 4349The algorithm mp\_count\_bits calculates the number of bits in an mp\_int which is used to find the initial value of $p$. The call to mp\_div\_2d 4350on line @31,mp_div_2d@ calculates both the quotient $q$ and the remainder $a$ required. By doing both in a single function call the code size 4351is kept fairly small. The multiplication by $k$ is only performed if $k > 1$. This allows reductions modulo $2^p - 1$ to be performed without 4352any multiplications. 4353 4354The unsigned s\_mp\_add, mp\_cmp\_mag and s\_mp\_sub are used in place of their full sign counterparts since the inputs are only valid if they are 4355positive. By using the unsigned versions the overhead is kept to a minimum. 4356 4357\subsubsection{Unrestricted Setup} 4358To setup this reduction algorithm the value of $k = 2^p - n$ is required. 4359 4360\begin{figure}[!here] 4361\begin{small} 4362\begin{center} 4363\begin{tabular}{l} 4364\hline Algorithm \textbf{mp\_reduce\_2k\_setup}. \\ 4365\textbf{Input}. mp\_int $n$ \\ 4366\textbf{Output}. $k = 2^p - n$ \\ 4367\hline 43681. $p \leftarrow \lceil lg(n) \rceil$ (\textit{mp\_count\_bits}) \\ 43692. $x \leftarrow 2^p$ (\textit{mp\_2expt}) \\ 43703. $x \leftarrow x - n$ (\textit{mp\_sub}) \\ 43714. $k \leftarrow x_0$ \\ 43725. Return(\textit{MP\_OKAY}). \\ 4373\hline 4374\end{tabular} 4375\end{center} 4376\end{small} 4377\caption{Algorithm mp\_reduce\_2k\_setup} 4378\end{figure} 4379 4380\textbf{Algorithm mp\_reduce\_2k\_setup.} 4381This algorithm computes the value of $k$ required for the algorithm mp\_reduce\_2k. By making a temporary variable $x$ equal to $2^p$ a subtraction 4382is sufficient to solve for $k$. Alternatively if $n$ has more than one digit the value of $k$ is simply $\beta - n_0$. 4383 4384EXAM,bn_mp_reduce_2k_setup.c 4385 4386\subsubsection{Unrestricted Detection} 4387An integer $n$ is a valid unrestricted Diminished Radix modulus if either of the following are true. 4388 4389\begin{enumerate} 4390\item The number has only one digit. 4391\item The number has more than one digit and every bit from the $\beta$'th to the most significant is one. 4392\end{enumerate} 4393 4394If either condition is true than there is a power of two $2^p$ such that $0 < 2^p - n < \beta$. If the input is only 4395one digit than it will always be of the correct form. Otherwise all of the bits above the first digit must be one. This arises from the fact 4396that there will be value of $k$ that when added to the modulus causes a carry in the first digit which propagates all the way to the most 4397significant bit. The resulting sum will be a power of two. 4398 4399\begin{figure}[!here] 4400\begin{small} 4401\begin{center} 4402\begin{tabular}{l} 4403\hline Algorithm \textbf{mp\_reduce\_is\_2k}. \\ 4404\textbf{Input}. mp\_int $n$ \\ 4405\textbf{Output}. $1$ if of proper form, $0$ otherwise \\ 4406\hline 44071. If $n.used = 0$ then return($0$). \\ 44082. If $n.used = 1$ then return($1$). \\ 44093. $p \leftarrow \lceil lg(n) \rceil$ (\textit{mp\_count\_bits}) \\ 44104. for $x$ from $lg(\beta)$ to $p$ do \\ 4411\hspace{3mm}4.1 If the ($x \mbox{ mod }lg(\beta)$)'th bit of the $\lfloor x / lg(\beta) \rfloor$ of $n$ is zero then return($0$). \\ 44125. Return($1$). \\ 4413\hline 4414\end{tabular} 4415\end{center} 4416\end{small} 4417\caption{Algorithm mp\_reduce\_is\_2k} 4418\end{figure} 4419 4420\textbf{Algorithm mp\_reduce\_is\_2k.} 4421This algorithm quickly determines if a modulus is of the form required for algorithm mp\_reduce\_2k to function properly. 4422 4423EXAM,bn_mp_reduce_is_2k.c 4424 4425 4426 4427\section{Algorithm Comparison} 4428So far three very different algorithms for modular reduction have been discussed. Each of the algorithms have their own strengths and weaknesses 4429that makes having such a selection very useful. The following table sumarizes the three algorithms along with comparisons of work factors. Since 4430all three algorithms have the restriction that $0 \le x < n^2$ and $n > 1$ those limitations are not included in the table. 4431 4432\begin{center} 4433\begin{small} 4434\begin{tabular}{|c|c|c|c|c|c|} 4435\hline \textbf{Method} & \textbf{Work Required} & \textbf{Limitations} & \textbf{$m = 8$} & \textbf{$m = 32$} & \textbf{$m = 64$} \\ 4436\hline Barrett & $m^2 + 2m - 1$ & None & $79$ & $1087$ & $4223$ \\ 4437\hline Montgomery & $m^2 + m$ & $n$ must be odd & $72$ & $1056$ & $4160$ \\ 4438\hline D.R. & $2m$ & $n = \beta^m - k$ & $16$ & $64$ & $128$ \\ 4439\hline 4440\end{tabular} 4441\end{small} 4442\end{center} 4443 4444In theory Montgomery and Barrett reductions would require roughly the same amount of time to complete. However, in practice since Montgomery 4445reduction can be written as a single function with the Comba technique it is much faster. Barrett reduction suffers from the overhead of 4446calling the half precision multipliers, addition and division by $\beta$ algorithms. 4447 4448For almost every cryptographic algorithm Montgomery reduction is the algorithm of choice. The one set of algorithms where Diminished Radix reduction truly 4449shines are based on the discrete logarithm problem such as Diffie-Hellman \cite{DH} and ElGamal \cite{ELGAMAL}. In these algorithms 4450primes of the form $\beta^m - k$ can be found and shared amongst users. These primes will allow the Diminished Radix algorithm to be used in 4451modular exponentiation to greatly speed up the operation. 4452 4453 4454 4455\section*{Exercises} 4456\begin{tabular}{cl} 4457$\left [ 3 \right ]$ & Prove that the ``trick'' in algorithm mp\_montgomery\_setup actually \\ 4458 & calculates the correct value of $\rho$. \\ 4459 & \\ 4460$\left [ 2 \right ]$ & Devise an algorithm to reduce modulo $n + k$ for small $k$ quickly. \\ 4461 & \\ 4462$\left [ 4 \right ]$ & Prove that the pseudo-code algorithm ``Diminished Radix Reduction'' \\ 4463 & (\textit{figure~\ref{fig:DR}}) terminates. Also prove the probability that it will \\ 4464 & terminate within $1 \le k \le 10$ iterations. \\ 4465 & \\ 4466\end{tabular} 4467 4468 4469\chapter{Exponentiation} 4470Exponentiation is the operation of raising one variable to the power of another, for example, $a^b$. A variant of exponentiation, computed 4471in a finite field or ring, is called modular exponentiation. This latter style of operation is typically used in public key 4472cryptosystems such as RSA and Diffie-Hellman. The ability to quickly compute modular exponentiations is of great benefit to any 4473such cryptosystem and many methods have been sought to speed it up. 4474 4475\section{Exponentiation Basics} 4476A trivial algorithm would simply multiply $a$ against itself $b - 1$ times to compute the exponentiation desired. However, as $b$ grows in size 4477the number of multiplications becomes prohibitive. Imagine what would happen if $b$ $\approx$ $2^{1024}$ as is the case when computing an RSA signature 4478with a $1024$-bit key. Such a calculation could never be completed as it would take simply far too long. 4479 4480Fortunately there is a very simple algorithm based on the laws of exponents. Recall that $lg_a(a^b) = b$ and that $lg_a(a^ba^c) = b + c$ which 4481are two trivial relationships between the base and the exponent. Let $b_i$ represent the $i$'th bit of $b$ starting from the least 4482significant bit. If $b$ is a $k$-bit integer than the following equation is true. 4483 4484\begin{equation} 4485a^b = \prod_{i=0}^{k-1} a^{2^i \cdot b_i} 4486\end{equation} 4487 4488By taking the base $a$ logarithm of both sides of the equation the following equation is the result. 4489 4490\begin{equation} 4491b = \sum_{i=0}^{k-1}2^i \cdot b_i 4492\end{equation} 4493 4494The term $a^{2^i}$ can be found from the $i - 1$'th term by squaring the term since $\left ( a^{2^i} \right )^2$ is equal to 4495$a^{2^{i+1}}$. This observation forms the basis of essentially all fast exponentiation algorithms. It requires $k$ squarings and on average 4496$k \over 2$ multiplications to compute the result. This is indeed quite an improvement over simply multiplying by $a$ a total of $b-1$ times. 4497 4498While this current method is a considerable speed up there are further improvements to be made. For example, the $a^{2^i}$ term does not need to 4499be computed in an auxilary variable. Consider the following equivalent algorithm. 4500 4501\begin{figure}[!here] 4502\begin{small} 4503\begin{center} 4504\begin{tabular}{l} 4505\hline Algorithm \textbf{Left to Right Exponentiation}. \\ 4506\textbf{Input}. Integer $a$, $b$ and $k$ \\ 4507\textbf{Output}. $c = a^b$ \\ 4508\hline \\ 45091. $c \leftarrow 1$ \\ 45102. for $i$ from $k - 1$ to $0$ do \\ 4511\hspace{3mm}2.1 $c \leftarrow c^2$ \\ 4512\hspace{3mm}2.2 $c \leftarrow c \cdot a^{b_i}$ \\ 45133. Return $c$. \\ 4514\hline 4515\end{tabular} 4516\end{center} 4517\end{small} 4518\caption{Left to Right Exponentiation} 4519\label{fig:LTOR} 4520\end{figure} 4521 4522This algorithm starts from the most significant bit and works towards the least significant bit. When the $i$'th bit of $b$ is set $a$ is 4523multiplied against the current product. In each iteration the product is squared which doubles the exponent of the individual terms of the 4524product. 4525 4526For example, let $b = 101100_2 \equiv 44_{10}$. The following chart demonstrates the actions of the algorithm. 4527 4528\newpage\begin{figure} 4529\begin{center} 4530\begin{tabular}{|c|c|} 4531\hline \textbf{Value of $i$} & \textbf{Value of $c$} \\ 4532\hline - & $1$ \\ 4533\hline $5$ & $a$ \\ 4534\hline $4$ & $a^2$ \\ 4535\hline $3$ & $a^4 \cdot a$ \\ 4536\hline $2$ & $a^8 \cdot a^2 \cdot a$ \\ 4537\hline $1$ & $a^{16} \cdot a^4 \cdot a^2$ \\ 4538\hline $0$ & $a^{32} \cdot a^8 \cdot a^4$ \\ 4539\hline 4540\end{tabular} 4541\end{center} 4542\caption{Example of Left to Right Exponentiation} 4543\end{figure} 4544 4545When the product $a^{32} \cdot a^8 \cdot a^4$ is simplified it is equal $a^{44}$ which is the desired exponentiation. This particular algorithm is 4546called ``Left to Right'' because it reads the exponent in that order. All of the exponentiation algorithms that will be presented are of this nature. 4547 4548\subsection{Single Digit Exponentiation} 4549The first algorithm in the series of exponentiation algorithms will be an unbounded algorithm where the exponent is a single digit. It is intended 4550to be used when a small power of an input is required (\textit{e.g. $a^5$}). It is faster than simply multiplying $b - 1$ times for all values of 4551$b$ that are greater than three. 4552 4553\newpage\begin{figure}[!here] 4554\begin{small} 4555\begin{center} 4556\begin{tabular}{l} 4557\hline Algorithm \textbf{mp\_expt\_d}. \\ 4558\textbf{Input}. mp\_int $a$ and mp\_digit $b$ \\ 4559\textbf{Output}. $c = a^b$ \\ 4560\hline \\ 45611. $g \leftarrow a$ (\textit{mp\_init\_copy}) \\ 45622. $c \leftarrow 1$ (\textit{mp\_set}) \\ 45633. for $x$ from 1 to $lg(\beta)$ do \\ 4564\hspace{3mm}3.1 $c \leftarrow c^2$ (\textit{mp\_sqr}) \\ 4565\hspace{3mm}3.2 If $b$ AND $2^{lg(\beta) - 1} \ne 0$ then \\ 4566\hspace{6mm}3.2.1 $c \leftarrow c \cdot g$ (\textit{mp\_mul}) \\ 4567\hspace{3mm}3.3 $b \leftarrow b << 1$ \\ 45684. Clear $g$. \\ 45695. Return(\textit{MP\_OKAY}). \\ 4570\hline 4571\end{tabular} 4572\end{center} 4573\end{small} 4574\caption{Algorithm mp\_expt\_d} 4575\end{figure} 4576 4577\textbf{Algorithm mp\_expt\_d.} 4578This algorithm computes the value of $a$ raised to the power of a single digit $b$. It uses the left to right exponentiation algorithm to 4579quickly compute the exponentiation. It is loosely based on algorithm 14.79 of HAC \cite[pp. 615]{HAC} with the difference that the 4580exponent is a fixed width. 4581 4582A copy of $a$ is made first to allow destination variable $c$ be the same as the source variable $a$. The result is set to the initial value of 4583$1$ in the subsequent step. 4584 4585Inside the loop the exponent is read from the most significant bit first down to the least significant bit. First $c$ is invariably squared 4586on step 3.1. In the following step if the most significant bit of $b$ is one the copy of $a$ is multiplied against $c$. The value 4587of $b$ is shifted left one bit to make the next bit down from the most signficant bit the new most significant bit. In effect each 4588iteration of the loop moves the bits of the exponent $b$ upwards to the most significant location. 4589 4590EXAM,bn_mp_expt_d.c 4591 4592Line @29,mp_set@ sets the initial value of the result to $1$. Next the loop on line @31,for@ steps through each bit of the exponent starting from 4593the most significant down towards the least significant. The invariant squaring operation placed on line @333,mp_sqr@ is performed first. After 4594the squaring the result $c$ is multiplied by the base $g$ if and only if the most significant bit of the exponent is set. The shift on line 4595@47,<<@ moves all of the bits of the exponent upwards towards the most significant location. 4596 4597\section{$k$-ary Exponentiation} 4598When calculating an exponentiation the most time consuming bottleneck is the multiplications which are in general a small factor 4599slower than squaring. Recall from the previous algorithm that $b_{i}$ refers to the $i$'th bit of the exponent $b$. Suppose instead it referred to 4600the $i$'th $k$-bit digit of the exponent of $b$. For $k = 1$ the definitions are synonymous and for $k > 1$ algorithm~\ref{fig:KARY} 4601computes the same exponentiation. A group of $k$ bits from the exponent is called a \textit{window}. That is it is a small window on only a 4602portion of the entire exponent. Consider the following modification to the basic left to right exponentiation algorithm. 4603 4604\begin{figure}[!here] 4605\begin{small} 4606\begin{center} 4607\begin{tabular}{l} 4608\hline Algorithm \textbf{$k$-ary Exponentiation}. \\ 4609\textbf{Input}. Integer $a$, $b$, $k$ and $t$ \\ 4610\textbf{Output}. $c = a^b$ \\ 4611\hline \\ 46121. $c \leftarrow 1$ \\ 46132. for $i$ from $t - 1$ to $0$ do \\ 4614\hspace{3mm}2.1 $c \leftarrow c^{2^k} $ \\ 4615\hspace{3mm}2.2 Extract the $i$'th $k$-bit word from $b$ and store it in $g$. \\ 4616\hspace{3mm}2.3 $c \leftarrow c \cdot a^g$ \\ 46173. Return $c$. \\ 4618\hline 4619\end{tabular} 4620\end{center} 4621\end{small} 4622\caption{$k$-ary Exponentiation} 4623\label{fig:KARY} 4624\end{figure} 4625 4626The squaring on step 2.1 can be calculated by squaring the value $c$ successively $k$ times. If the values of $a^g$ for $0 < g < 2^k$ have been 4627precomputed this algorithm requires only $t$ multiplications and $tk$ squarings. The table can be generated with $2^{k - 1} - 1$ squarings and 4628$2^{k - 1} + 1$ multiplications. This algorithm assumes that the number of bits in the exponent is evenly divisible by $k$. 4629However, when it is not the remaining $0 < x \le k - 1$ bits can be handled with algorithm~\ref{fig:LTOR}. 4630 4631Suppose $k = 4$ and $t = 100$. This modified algorithm will require $109$ multiplications and $408$ squarings to compute the exponentiation. The 4632original algorithm would on average have required $200$ multiplications and $400$ squrings to compute the same value. The total number of squarings 4633has increased slightly but the number of multiplications has nearly halved. 4634 4635\subsection{Optimal Values of $k$} 4636An optimal value of $k$ will minimize $2^{k} + \lceil n / k \rceil + n - 1$ for a fixed number of bits in the exponent $n$. The simplest 4637approach is to brute force search amongst the values $k = 2, 3, \ldots, 8$ for the lowest result. Table~\ref{fig:OPTK} lists optimal values of $k$ 4638for various exponent sizes and compares the number of multiplication and squarings required against algorithm~\ref{fig:LTOR}. 4639 4640\begin{figure}[here] 4641\begin{center} 4642\begin{small} 4643\begin{tabular}{|c|c|c|c|c|c|} 4644\hline \textbf{Exponent (bits)} & \textbf{Optimal $k$} & \textbf{Work at $k$} & \textbf{Work with ~\ref{fig:LTOR}} \\ 4645\hline $16$ & $2$ & $27$ & $24$ \\ 4646\hline $32$ & $3$ & $49$ & $48$ \\ 4647\hline $64$ & $3$ & $92$ & $96$ \\ 4648\hline $128$ & $4$ & $175$ & $192$ \\ 4649\hline $256$ & $4$ & $335$ & $384$ \\ 4650\hline $512$ & $5$ & $645$ & $768$ \\ 4651\hline $1024$ & $6$ & $1257$ & $1536$ \\ 4652\hline $2048$ & $6$ & $2452$ & $3072$ \\ 4653\hline $4096$ & $7$ & $4808$ & $6144$ \\ 4654\hline 4655\end{tabular} 4656\end{small} 4657\end{center} 4658\caption{Optimal Values of $k$ for $k$-ary Exponentiation} 4659\label{fig:OPTK} 4660\end{figure} 4661 4662\subsection{Sliding-Window Exponentiation} 4663A simple modification to the previous algorithm is only generate the upper half of the table in the range $2^{k-1} \le g < 2^k$. Essentially 4664this is a table for all values of $g$ where the most significant bit of $g$ is a one. However, in order for this to be allowed in the 4665algorithm values of $g$ in the range $0 \le g < 2^{k-1}$ must be avoided. 4666 4667Table~\ref{fig:OPTK2} lists optimal values of $k$ for various exponent sizes and compares the work required against algorithm~\ref{fig:KARY}. 4668 4669\begin{figure}[here] 4670\begin{center} 4671\begin{small} 4672\begin{tabular}{|c|c|c|c|c|c|} 4673\hline \textbf{Exponent (bits)} & \textbf{Optimal $k$} & \textbf{Work at $k$} & \textbf{Work with ~\ref{fig:KARY}} \\ 4674\hline $16$ & $3$ & $24$ & $27$ \\ 4675\hline $32$ & $3$ & $45$ & $49$ \\ 4676\hline $64$ & $4$ & $87$ & $92$ \\ 4677\hline $128$ & $4$ & $167$ & $175$ \\ 4678\hline $256$ & $5$ & $322$ & $335$ \\ 4679\hline $512$ & $6$ & $628$ & $645$ \\ 4680\hline $1024$ & $6$ & $1225$ & $1257$ \\ 4681\hline $2048$ & $7$ & $2403$ & $2452$ \\ 4682\hline $4096$ & $8$ & $4735$ & $4808$ \\ 4683\hline 4684\end{tabular} 4685\end{small} 4686\end{center} 4687\caption{Optimal Values of $k$ for Sliding Window Exponentiation} 4688\label{fig:OPTK2} 4689\end{figure} 4690 4691\newpage\begin{figure}[!here] 4692\begin{small} 4693\begin{center} 4694\begin{tabular}{l} 4695\hline Algorithm \textbf{Sliding Window $k$-ary Exponentiation}. \\ 4696\textbf{Input}. Integer $a$, $b$, $k$ and $t$ \\ 4697\textbf{Output}. $c = a^b$ \\ 4698\hline \\ 46991. $c \leftarrow 1$ \\ 47002. for $i$ from $t - 1$ to $0$ do \\ 4701\hspace{3mm}2.1 If the $i$'th bit of $b$ is a zero then \\ 4702\hspace{6mm}2.1.1 $c \leftarrow c^2$ \\ 4703\hspace{3mm}2.2 else do \\ 4704\hspace{6mm}2.2.1 $c \leftarrow c^{2^k}$ \\ 4705\hspace{6mm}2.2.2 Extract the $k$ bits from $(b_{i}b_{i-1}\ldots b_{i-(k-1)})$ and store it in $g$. \\ 4706\hspace{6mm}2.2.3 $c \leftarrow c \cdot a^g$ \\ 4707\hspace{6mm}2.2.4 $i \leftarrow i - k$ \\ 47083. Return $c$. \\ 4709\hline 4710\end{tabular} 4711\end{center} 4712\end{small} 4713\caption{Sliding Window $k$-ary Exponentiation} 4714\end{figure} 4715 4716Similar to the previous algorithm this algorithm must have a special handler when fewer than $k$ bits are left in the exponent. While this 4717algorithm requires the same number of squarings it can potentially have fewer multiplications. The pre-computed table $a^g$ is also half 4718the size as the previous table. 4719 4720Consider the exponent $b = 111101011001000_2 \equiv 31432_{10}$ with $k = 3$ using both algorithms. The first algorithm will divide the exponent up as 4721the following five $3$-bit words $b \equiv \left ( 111, 101, 011, 001, 000 \right )_{2}$. The second algorithm will break the 4722exponent as $b \equiv \left ( 111, 101, 0, 110, 0, 100, 0 \right )_{2}$. The single digit $0$ in the second representation are where 4723a single squaring took place instead of a squaring and multiplication. In total the first method requires $10$ multiplications and $18$ 4724squarings. The second method requires $8$ multiplications and $18$ squarings. 4725 4726In general the sliding window method is never slower than the generic $k$-ary method and often it is slightly faster. 4727 4728\section{Modular Exponentiation} 4729 4730Modular exponentiation is essentially computing the power of a base within a finite field or ring. For example, computing 4731$d \equiv a^b \mbox{ (mod }c\mbox{)}$ is a modular exponentiation. Instead of first computing $a^b$ and then reducing it 4732modulo $c$ the intermediate result is reduced modulo $c$ after every squaring or multiplication operation. 4733 4734This guarantees that any intermediate result is bounded by $0 \le d \le c^2 - 2c + 1$ and can be reduced modulo $c$ quickly using 4735one of the algorithms presented in ~REDUCTION~. 4736 4737Before the actual modular exponentiation algorithm can be written a wrapper algorithm must be written first. This algorithm 4738will allow the exponent $b$ to be negative which is computed as $c \equiv \left (1 / a \right )^{\vert b \vert} \mbox{(mod }d\mbox{)}$. The 4739value of $(1/a) \mbox{ mod }c$ is computed using the modular inverse (\textit{see \ref{sec;modinv}}). If no inverse exists the algorithm 4740terminates with an error. 4741 4742\begin{figure}[!here] 4743\begin{small} 4744\begin{center} 4745\begin{tabular}{l} 4746\hline Algorithm \textbf{mp\_exptmod}. \\ 4747\textbf{Input}. mp\_int $a$, $b$ and $c$ \\ 4748\textbf{Output}. $y \equiv g^x \mbox{ (mod }p\mbox{)}$ \\ 4749\hline \\ 47501. If $c.sign = MP\_NEG$ return(\textit{MP\_VAL}). \\ 47512. If $b.sign = MP\_NEG$ then \\ 4752\hspace{3mm}2.1 $g' \leftarrow g^{-1} \mbox{ (mod }c\mbox{)}$ \\ 4753\hspace{3mm}2.2 $x' \leftarrow \vert x \vert$ \\ 4754\hspace{3mm}2.3 Compute $d \equiv g'^{x'} \mbox{ (mod }c\mbox{)}$ via recursion. \\ 47553. if $p$ is odd \textbf{OR} $p$ is a D.R. modulus then \\ 4756\hspace{3mm}3.1 Compute $y \equiv g^{x} \mbox{ (mod }p\mbox{)}$ via algorithm mp\_exptmod\_fast. \\ 47574. else \\ 4758\hspace{3mm}4.1 Compute $y \equiv g^{x} \mbox{ (mod }p\mbox{)}$ via algorithm s\_mp\_exptmod. \\ 4759\hline 4760\end{tabular} 4761\end{center} 4762\end{small} 4763\caption{Algorithm mp\_exptmod} 4764\end{figure} 4765 4766\textbf{Algorithm mp\_exptmod.} 4767The first algorithm which actually performs modular exponentiation is algorithm s\_mp\_exptmod. It is a sliding window $k$-ary algorithm 4768which uses Barrett reduction to reduce the product modulo $p$. The second algorithm mp\_exptmod\_fast performs the same operation 4769except it uses either Montgomery or Diminished Radix reduction. The two latter reduction algorithms are clumped in the same exponentiation 4770algorithm since their arguments are essentially the same (\textit{two mp\_ints and one mp\_digit}). 4771 4772EXAM,bn_mp_exptmod.c 4773 4774In order to keep the algorithms in a known state the first step on line @29,if@ is to reject any negative modulus as input. If the exponent is 4775negative the algorithm tries to perform a modular exponentiation with the modular inverse of the base $G$. The temporary variable $tmpG$ is assigned 4776the modular inverse of $G$ and $tmpX$ is assigned the absolute value of $X$. The algorithm will recuse with these new values with a positive 4777exponent. 4778 4779If the exponent is positive the algorithm resumes the exponentiation. Line @63,dr_@ determines if the modulus is of the restricted Diminished Radix 4780form. If it is not line @65,reduce@ attempts to determine if it is of a unrestricted Diminished Radix form. The integer $dr$ will take on one 4781of three values. 4782 4783\begin{enumerate} 4784\item $dr = 0$ means that the modulus is not of either restricted or unrestricted Diminished Radix form. 4785\item $dr = 1$ means that the modulus is of restricted Diminished Radix form. 4786\item $dr = 2$ means that the modulus is of unrestricted Diminished Radix form. 4787\end{enumerate} 4788 4789Line @69,if@ determines if the fast modular exponentiation algorithm can be used. It is allowed if $dr \ne 0$ or if the modulus is odd. Otherwise, 4790the slower s\_mp\_exptmod algorithm is used which uses Barrett reduction. 4791 4792\subsection{Barrett Modular Exponentiation} 4793 4794\newpage\begin{figure}[!here] 4795\begin{small} 4796\begin{center} 4797\begin{tabular}{l} 4798\hline Algorithm \textbf{s\_mp\_exptmod}. \\ 4799\textbf{Input}. mp\_int $a$, $b$ and $c$ \\ 4800\textbf{Output}. $y \equiv g^x \mbox{ (mod }p\mbox{)}$ \\ 4801\hline \\ 48021. $k \leftarrow lg(x)$ \\ 48032. $winsize \leftarrow \left \lbrace \begin{array}{ll} 4804 2 & \mbox{if }k \le 7 \\ 4805 3 & \mbox{if }7 < k \le 36 \\ 4806 4 & \mbox{if }36 < k \le 140 \\ 4807 5 & \mbox{if }140 < k \le 450 \\ 4808 6 & \mbox{if }450 < k \le 1303 \\ 4809 7 & \mbox{if }1303 < k \le 3529 \\ 4810 8 & \mbox{if }3529 < k \\ 4811 \end{array} \right .$ \\ 48123. Initialize $2^{winsize}$ mp\_ints in an array named $M$ and one mp\_int named $\mu$ \\ 48134. Calculate the $\mu$ required for Barrett Reduction (\textit{mp\_reduce\_setup}). \\ 48145. $M_1 \leftarrow g \mbox{ (mod }p\mbox{)}$ \\ 4815\\ 4816Setup the table of small powers of $g$. First find $g^{2^{winsize}}$ and then all multiples of it. \\ 48176. $k \leftarrow 2^{winsize - 1}$ \\ 48187. $M_{k} \leftarrow M_1$ \\ 48198. for $ix$ from 0 to $winsize - 2$ do \\ 4820\hspace{3mm}8.1 $M_k \leftarrow \left ( M_k \right )^2$ (\textit{mp\_sqr}) \\ 4821\hspace{3mm}8.2 $M_k \leftarrow M_k \mbox{ (mod }p\mbox{)}$ (\textit{mp\_reduce}) \\ 48229. for $ix$ from $2^{winsize - 1} + 1$ to $2^{winsize} - 1$ do \\ 4823\hspace{3mm}9.1 $M_{ix} \leftarrow M_{ix - 1} \cdot M_{1}$ (\textit{mp\_mul}) \\ 4824\hspace{3mm}9.2 $M_{ix} \leftarrow M_{ix} \mbox{ (mod }p\mbox{)}$ (\textit{mp\_reduce}) \\ 482510. $res \leftarrow 1$ \\ 4826\\ 4827Start Sliding Window. \\ 482811. $mode \leftarrow 0, bitcnt \leftarrow 1, buf \leftarrow 0, digidx \leftarrow x.used - 1, bitcpy \leftarrow 0, bitbuf \leftarrow 0$ \\ 482912. Loop \\ 4830\hspace{3mm}12.1 $bitcnt \leftarrow bitcnt - 1$ \\ 4831\hspace{3mm}12.2 If $bitcnt = 0$ then do \\ 4832\hspace{6mm}12.2.1 If $digidx = -1$ goto step 13. \\ 4833\hspace{6mm}12.2.2 $buf \leftarrow x_{digidx}$ \\ 4834\hspace{6mm}12.2.3 $digidx \leftarrow digidx - 1$ \\ 4835\hspace{6mm}12.2.4 $bitcnt \leftarrow lg(\beta)$ \\ 4836Continued on next page. \\ 4837\hline 4838\end{tabular} 4839\end{center} 4840\end{small} 4841\caption{Algorithm s\_mp\_exptmod} 4842\end{figure} 4843 4844\newpage\begin{figure}[!here] 4845\begin{small} 4846\begin{center} 4847\begin{tabular}{l} 4848\hline Algorithm \textbf{s\_mp\_exptmod} (\textit{continued}). \\ 4849\textbf{Input}. mp\_int $a$, $b$ and $c$ \\ 4850\textbf{Output}. $y \equiv g^x \mbox{ (mod }p\mbox{)}$ \\ 4851\hline \\ 4852\hspace{3mm}12.3 $y \leftarrow (buf >> (lg(\beta) - 1))$ AND $1$ \\ 4853\hspace{3mm}12.4 $buf \leftarrow buf << 1$ \\ 4854\hspace{3mm}12.5 if $mode = 0$ and $y = 0$ then goto step 12. \\ 4855\hspace{3mm}12.6 if $mode = 1$ and $y = 0$ then do \\ 4856\hspace{6mm}12.6.1 $res \leftarrow res^2$ \\ 4857\hspace{6mm}12.6.2 $res \leftarrow res \mbox{ (mod }p\mbox{)}$ \\ 4858\hspace{6mm}12.6.3 Goto step 12. \\ 4859\hspace{3mm}12.7 $bitcpy \leftarrow bitcpy + 1$ \\ 4860\hspace{3mm}12.8 $bitbuf \leftarrow bitbuf + (y << (winsize - bitcpy))$ \\ 4861\hspace{3mm}12.9 $mode \leftarrow 2$ \\ 4862\hspace{3mm}12.10 If $bitcpy = winsize$ then do \\ 4863\hspace{6mm}Window is full so perform the squarings and single multiplication. \\ 4864\hspace{6mm}12.10.1 for $ix$ from $0$ to $winsize -1$ do \\ 4865\hspace{9mm}12.10.1.1 $res \leftarrow res^2$ \\ 4866\hspace{9mm}12.10.1.2 $res \leftarrow res \mbox{ (mod }p\mbox{)}$ \\ 4867\hspace{6mm}12.10.2 $res \leftarrow res \cdot M_{bitbuf}$ \\ 4868\hspace{6mm}12.10.3 $res \leftarrow res \mbox{ (mod }p\mbox{)}$ \\ 4869\hspace{6mm}Reset the window. \\ 4870\hspace{6mm}12.10.4 $bitcpy \leftarrow 0, bitbuf \leftarrow 0, mode \leftarrow 1$ \\ 4871\\ 4872No more windows left. Check for residual bits of exponent. \\ 487313. If $mode = 2$ and $bitcpy > 0$ then do \\ 4874\hspace{3mm}13.1 for $ix$ form $0$ to $bitcpy - 1$ do \\ 4875\hspace{6mm}13.1.1 $res \leftarrow res^2$ \\ 4876\hspace{6mm}13.1.2 $res \leftarrow res \mbox{ (mod }p\mbox{)}$ \\ 4877\hspace{6mm}13.1.3 $bitbuf \leftarrow bitbuf << 1$ \\ 4878\hspace{6mm}13.1.4 If $bitbuf$ AND $2^{winsize} \ne 0$ then do \\ 4879\hspace{9mm}13.1.4.1 $res \leftarrow res \cdot M_{1}$ \\ 4880\hspace{9mm}13.1.4.2 $res \leftarrow res \mbox{ (mod }p\mbox{)}$ \\ 488114. $y \leftarrow res$ \\ 488215. Clear $res$, $mu$ and the $M$ array. \\ 488316. Return(\textit{MP\_OKAY}). \\ 4884\hline 4885\end{tabular} 4886\end{center} 4887\end{small} 4888\caption{Algorithm s\_mp\_exptmod (continued)} 4889\end{figure} 4890 4891\textbf{Algorithm s\_mp\_exptmod.} 4892This algorithm computes the $x$'th power of $g$ modulo $p$ and stores the result in $y$. It takes advantage of the Barrett reduction 4893algorithm to keep the product small throughout the algorithm. 4894 4895The first two steps determine the optimal window size based on the number of bits in the exponent. The larger the exponent the 4896larger the window size becomes. After a window size $winsize$ has been chosen an array of $2^{winsize}$ mp\_int variables is allocated. This 4897table will hold the values of $g^x \mbox{ (mod }p\mbox{)}$ for $2^{winsize - 1} \le x < 2^{winsize}$. 4898 4899After the table is allocated the first power of $g$ is found. Since $g \ge p$ is allowed it must be first reduced modulo $p$ to make 4900the rest of the algorithm more efficient. The first element of the table at $2^{winsize - 1}$ is found by squaring $M_1$ successively $winsize - 2$ 4901times. The rest of the table elements are found by multiplying the previous element by $M_1$ modulo $p$. 4902 4903Now that the table is available the sliding window may begin. The following list describes the functions of all the variables in the window. 4904\begin{enumerate} 4905\item The variable $mode$ dictates how the bits of the exponent are interpreted. 4906\begin{enumerate} 4907 \item When $mode = 0$ the bits are ignored since no non-zero bit of the exponent has been seen yet. For example, if the exponent were simply 4908 $1$ then there would be $lg(\beta) - 1$ zero bits before the first non-zero bit. In this case bits are ignored until a non-zero bit is found. 4909 \item When $mode = 1$ a non-zero bit has been seen before and a new $winsize$-bit window has not been formed yet. In this mode leading $0$ bits 4910 are read and a single squaring is performed. If a non-zero bit is read a new window is created. 4911 \item When $mode = 2$ the algorithm is in the middle of forming a window and new bits are appended to the window from the most significant bit 4912 downwards. 4913\end{enumerate} 4914\item The variable $bitcnt$ indicates how many bits are left in the current digit of the exponent left to be read. When it reaches zero a new digit 4915 is fetched from the exponent. 4916\item The variable $buf$ holds the currently read digit of the exponent. 4917\item The variable $digidx$ is an index into the exponents digits. It starts at the leading digit $x.used - 1$ and moves towards the trailing digit. 4918\item The variable $bitcpy$ indicates how many bits are in the currently formed window. When it reaches $winsize$ the window is flushed and 4919 the appropriate operations performed. 4920\item The variable $bitbuf$ holds the current bits of the window being formed. 4921\end{enumerate} 4922 4923All of step 12 is the window processing loop. It will iterate while there are digits available form the exponent to read. The first step 4924inside this loop is to extract a new digit if no more bits are available in the current digit. If there are no bits left a new digit is 4925read and if there are no digits left than the loop terminates. 4926 4927After a digit is made available step 12.3 will extract the most significant bit of the current digit and move all other bits in the digit 4928upwards. In effect the digit is read from most significant bit to least significant bit and since the digits are read from leading to 4929trailing edges the entire exponent is read from most significant bit to least significant bit. 4930 4931At step 12.5 if the $mode$ and currently extracted bit $y$ are both zero the bit is ignored and the next bit is read. This prevents the 4932algorithm from having to perform trivial squaring and reduction operations before the first non-zero bit is read. Step 12.6 and 12.7-10 handle 4933the two cases of $mode = 1$ and $mode = 2$ respectively. 4934 4935FIGU,expt_state,Sliding Window State Diagram 4936 4937By step 13 there are no more digits left in the exponent. However, there may be partial bits in the window left. If $mode = 2$ then 4938a Left-to-Right algorithm is used to process the remaining few bits. 4939 4940EXAM,bn_s_mp_exptmod.c 4941 4942Lines @31,if@ through @45,}@ determine the optimal window size based on the length of the exponent in bits. The window divisions are sorted 4943from smallest to greatest so that in each \textbf{if} statement only one condition must be tested. For example, by the \textbf{if} statement 4944on line @37,if@ the value of $x$ is already known to be greater than $140$. 4945 4946The conditional piece of code beginning on line @42,ifdef@ allows the window size to be restricted to five bits. This logic is used to ensure 4947the table of precomputed powers of $G$ remains relatively small. 4948 4949The for loop on line @60,for@ initializes the $M$ array while lines @71,mp_init@ and @75,mp_reduce@ through @85,}@ initialize the reduction 4950function that will be used for this modulus. 4951 4952-- More later. 4953 4954\section{Quick Power of Two} 4955Calculating $b = 2^a$ can be performed much quicker than with any of the previous algorithms. Recall that a logical shift left $m << k$ is 4956equivalent to $m \cdot 2^k$. By this logic when $m = 1$ a quick power of two can be achieved. 4957 4958\begin{figure}[!here] 4959\begin{small} 4960\begin{center} 4961\begin{tabular}{l} 4962\hline Algorithm \textbf{mp\_2expt}. \\ 4963\textbf{Input}. integer $b$ \\ 4964\textbf{Output}. $a \leftarrow 2^b$ \\ 4965\hline \\ 49661. $a \leftarrow 0$ \\ 49672. If $a.alloc < \lfloor b / lg(\beta) \rfloor + 1$ then grow $a$ appropriately. \\ 49683. $a.used \leftarrow \lfloor b / lg(\beta) \rfloor + 1$ \\ 49694. $a_{\lfloor b / lg(\beta) \rfloor} \leftarrow 1 << (b \mbox{ mod } lg(\beta))$ \\ 49705. Return(\textit{MP\_OKAY}). \\ 4971\hline 4972\end{tabular} 4973\end{center} 4974\end{small} 4975\caption{Algorithm mp\_2expt} 4976\end{figure} 4977 4978\textbf{Algorithm mp\_2expt.} 4979 4980EXAM,bn_mp_2expt.c 4981 4982\chapter{Higher Level Algorithms} 4983 4984This chapter discusses the various higher level algorithms that are required to complete a well rounded multiple precision integer package. These 4985routines are less performance oriented than the algorithms of chapters five, six and seven but are no less important. 4986 4987The first section describes a method of integer division with remainder that is universally well known. It provides the signed division logic 4988for the package. The subsequent section discusses a set of algorithms which allow a single digit to be the 2nd operand for a variety of operations. 4989These algorithms serve mostly to simplify other algorithms where small constants are required. The last two sections discuss how to manipulate 4990various representations of integers. For example, converting from an mp\_int to a string of character. 4991 4992\section{Integer Division with Remainder} 4993\label{sec:division} 4994 4995Integer division aside from modular exponentiation is the most intensive algorithm to compute. Like addition, subtraction and multiplication 4996the basis of this algorithm is the long-hand division algorithm taught to school children. Throughout this discussion several common variables 4997will be used. Let $x$ represent the divisor and $y$ represent the dividend. Let $q$ represent the integer quotient $\lfloor y / x \rfloor$ and 4998let $r$ represent the remainder $r = y - x \lfloor y / x \rfloor$. The following simple algorithm will be used to start the discussion. 4999 5000\newpage\begin{figure}[!here] 5001\begin{small} 5002\begin{center} 5003\begin{tabular}{l} 5004\hline Algorithm \textbf{Radix-$\beta$ Integer Division}. \\ 5005\textbf{Input}. integer $x$ and $y$ \\ 5006\textbf{Output}. $q = \lfloor y/x\rfloor, r = y - xq$ \\ 5007\hline \\ 50081. $q \leftarrow 0$ \\ 50092. $n \leftarrow \vert \vert y \vert \vert - \vert \vert x \vert \vert$ \\ 50103. for $t$ from $n$ down to $0$ do \\ 5011\hspace{3mm}3.1 Maximize $k$ such that $kx\beta^t$ is less than or equal to $y$ and $(k + 1)x\beta^t$ is greater. \\ 5012\hspace{3mm}3.2 $q \leftarrow q + k\beta^t$ \\ 5013\hspace{3mm}3.3 $y \leftarrow y - kx\beta^t$ \\ 50144. $r \leftarrow y$ \\ 50155. Return($q, r$) \\ 5016\hline 5017\end{tabular} 5018\end{center} 5019\end{small} 5020\caption{Algorithm Radix-$\beta$ Integer Division} 5021\label{fig:raddiv} 5022\end{figure} 5023 5024As children we are taught this very simple algorithm for the case of $\beta = 10$. Almost instinctively several optimizations are taught for which 5025their reason of existing are never explained. For this example let $y = 5471$ represent the dividend and $x = 23$ represent the divisor. 5026 5027To find the first digit of the quotient the value of $k$ must be maximized such that $kx\beta^t$ is less than or equal to $y$ and 5028simultaneously $(k + 1)x\beta^t$ is greater than $y$. Implicitly $k$ is the maximum value the $t$'th digit of the quotient may have. The habitual method 5029used to find the maximum is to ``eyeball'' the two numbers, typically only the leading digits and quickly estimate a quotient. By only using leading 5030digits a much simpler division may be used to form an educated guess at what the value must be. In this case $k = \lfloor 54/23\rfloor = 2$ quickly 5031arises as a possible solution. Indeed $2x\beta^2 = 4600$ is less than $y = 5471$ and simultaneously $(k + 1)x\beta^2 = 6900$ is larger than $y$. 5032As a result $k\beta^2$ is added to the quotient which now equals $q = 200$ and $4600$ is subtracted from $y$ to give a remainder of $y = 841$. 5033 5034Again this process is repeated to produce the quotient digit $k = 3$ which makes the quotient $q = 200 + 3\beta = 230$ and the remainder 5035$y = 841 - 3x\beta = 181$. Finally the last iteration of the loop produces $k = 7$ which leads to the quotient $q = 230 + 7 = 237$ and the 5036remainder $y = 181 - 7x = 20$. The final quotient and remainder found are $q = 237$ and $r = y = 20$ which are indeed correct since 5037$237 \cdot 23 + 20 = 5471$ is true. 5038 5039\subsection{Quotient Estimation} 5040\label{sec:divest} 5041As alluded to earlier the quotient digit $k$ can be estimated from only the leading digits of both the divisor and dividend. When $p$ leading 5042digits are used from both the divisor and dividend to form an estimation the accuracy of the estimation rises as $p$ grows. Technically 5043speaking the estimation is based on assuming the lower $\vert \vert y \vert \vert - p$ and $\vert \vert x \vert \vert - p$ lower digits of the 5044dividend and divisor are zero. 5045 5046The value of the estimation may off by a few values in either direction and in general is fairly correct. A simplification \cite[pp. 271]{TAOCPV2} 5047of the estimation technique is to use $t + 1$ digits of the dividend and $t$ digits of the divisor, in particularly when $t = 1$. The estimate 5048using this technique is never too small. For the following proof let $t = \vert \vert y \vert \vert - 1$ and $s = \vert \vert x \vert \vert - 1$ 5049represent the most significant digits of the dividend and divisor respectively. 5050 5051\textbf{Proof.}\textit{ The quotient $\hat k = \lfloor (y_t\beta + y_{t-1}) / x_s \rfloor$ is greater than or equal to 5052$k = \lfloor y / (x \cdot \beta^{\vert \vert y \vert \vert - \vert \vert x \vert \vert - 1}) \rfloor$. } 5053The first obvious case is when $\hat k = \beta - 1$ in which case the proof is concluded since the real quotient cannot be larger. For all other 5054cases $\hat k = \lfloor (y_t\beta + y_{t-1}) / x_s \rfloor$ and $\hat k x_s \ge y_t\beta + y_{t-1} - x_s + 1$. The latter portion of the inequalility 5055$-x_s + 1$ arises from the fact that a truncated integer division will give the same quotient for at most $x_s - 1$ values. Next a series of 5056inequalities will prove the hypothesis. 5057 5058\begin{equation} 5059y - \hat k x \le y - \hat k x_s\beta^s 5060\end{equation} 5061 5062This is trivially true since $x \ge x_s\beta^s$. Next we replace $\hat kx_s\beta^s$ by the previous inequality for $\hat kx_s$. 5063 5064\begin{equation} 5065y - \hat k x \le y_t\beta^t + \ldots + y_0 - (y_t\beta^t + y_{t-1}\beta^{t-1} - x_s\beta^t + \beta^s) 5066\end{equation} 5067 5068By simplifying the previous inequality the following inequality is formed. 5069 5070\begin{equation} 5071y - \hat k x \le y_{t-2}\beta^{t-2} + \ldots + y_0 + x_s\beta^s - \beta^s 5072\end{equation} 5073 5074Subsequently, 5075 5076\begin{equation} 5077y_{t-2}\beta^{t-2} + \ldots + y_0 + x_s\beta^s - \beta^s < x_s\beta^s \le x 5078\end{equation} 5079 5080Which proves that $y - \hat kx \le x$ and by consequence $\hat k \ge k$ which concludes the proof. \textbf{QED} 5081 5082 5083\subsection{Normalized Integers} 5084For the purposes of division a normalized input is when the divisors leading digit $x_n$ is greater than or equal to $\beta / 2$. By multiplying both 5085$x$ and $y$ by $j = \lfloor (\beta / 2) / x_n \rfloor$ the quotient remains unchanged and the remainder is simply $j$ times the original 5086remainder. The purpose of normalization is to ensure the leading digit of the divisor is sufficiently large such that the estimated quotient will 5087lie in the domain of a single digit. Consider the maximum dividend $(\beta - 1) \cdot \beta + (\beta - 1)$ and the minimum divisor $\beta / 2$. 5088 5089\begin{equation} 5090{{\beta^2 - 1} \over { \beta / 2}} \le 2\beta - {2 \over \beta} 5091\end{equation} 5092 5093At most the quotient approaches $2\beta$, however, in practice this will not occur since that would imply the previous quotient digit was too small. 5094 5095\subsection{Radix-$\beta$ Division with Remainder} 5096\newpage\begin{figure}[!here] 5097\begin{small} 5098\begin{center} 5099\begin{tabular}{l} 5100\hline Algorithm \textbf{mp\_div}. \\ 5101\textbf{Input}. mp\_int $a, b$ \\ 5102\textbf{Output}. $c = \lfloor a/b \rfloor$, $d = a - bc$ \\ 5103\hline \\ 51041. If $b = 0$ return(\textit{MP\_VAL}). \\ 51052. If $\vert a \vert < \vert b \vert$ then do \\ 5106\hspace{3mm}2.1 $d \leftarrow a$ \\ 5107\hspace{3mm}2.2 $c \leftarrow 0$ \\ 5108\hspace{3mm}2.3 Return(\textit{MP\_OKAY}). \\ 5109\\ 5110Setup the quotient to receive the digits. \\ 51113. Grow $q$ to $a.used + 2$ digits. \\ 51124. $q \leftarrow 0$ \\ 51135. $x \leftarrow \vert a \vert , y \leftarrow \vert b \vert$ \\ 51146. $sign \leftarrow \left \lbrace \begin{array}{ll} 5115 MP\_ZPOS & \mbox{if }a.sign = b.sign \\ 5116 MP\_NEG & \mbox{otherwise} \\ 5117 \end{array} \right .$ \\ 5118\\ 5119Normalize the inputs such that the leading digit of $y$ is greater than or equal to $\beta / 2$. \\ 51207. $norm \leftarrow (lg(\beta) - 1) - (\lceil lg(y) \rceil \mbox{ (mod }lg(\beta)\mbox{)})$ \\ 51218. $x \leftarrow x \cdot 2^{norm}, y \leftarrow y \cdot 2^{norm}$ \\ 5122\\ 5123Find the leading digit of the quotient. \\ 51249. $n \leftarrow x.used - 1, t \leftarrow y.used - 1$ \\ 512510. $y \leftarrow y \cdot \beta^{n - t}$ \\ 512611. While ($x \ge y$) do \\ 5127\hspace{3mm}11.1 $q_{n - t} \leftarrow q_{n - t} + 1$ \\ 5128\hspace{3mm}11.2 $x \leftarrow x - y$ \\ 512912. $y \leftarrow \lfloor y / \beta^{n-t} \rfloor$ \\ 5130\\ 5131Continued on the next page. \\ 5132\hline 5133\end{tabular} 5134\end{center} 5135\end{small} 5136\caption{Algorithm mp\_div} 5137\end{figure} 5138 5139\newpage\begin{figure}[!here] 5140\begin{small} 5141\begin{center} 5142\begin{tabular}{l} 5143\hline Algorithm \textbf{mp\_div} (continued). \\ 5144\textbf{Input}. mp\_int $a, b$ \\ 5145\textbf{Output}. $c = \lfloor a/b \rfloor$, $d = a - bc$ \\ 5146\hline \\ 5147Now find the remainder fo the digits. \\ 514813. for $i$ from $n$ down to $(t + 1)$ do \\ 5149\hspace{3mm}13.1 If $i > x.used$ then jump to the next iteration of this loop. \\ 5150\hspace{3mm}13.2 If $x_{i} = y_{t}$ then \\ 5151\hspace{6mm}13.2.1 $q_{i - t - 1} \leftarrow \beta - 1$ \\ 5152\hspace{3mm}13.3 else \\ 5153\hspace{6mm}13.3.1 $\hat r \leftarrow x_{i} \cdot \beta + x_{i - 1}$ \\ 5154\hspace{6mm}13.3.2 $\hat r \leftarrow \lfloor \hat r / y_{t} \rfloor$ \\ 5155\hspace{6mm}13.3.3 $q_{i - t - 1} \leftarrow \hat r$ \\ 5156\hspace{3mm}13.4 $q_{i - t - 1} \leftarrow q_{i - t - 1} + 1$ \\ 5157\\ 5158Fixup quotient estimation. \\ 5159\hspace{3mm}13.5 Loop \\ 5160\hspace{6mm}13.5.1 $q_{i - t - 1} \leftarrow q_{i - t - 1} - 1$ \\ 5161\hspace{6mm}13.5.2 t$1 \leftarrow 0$ \\ 5162\hspace{6mm}13.5.3 t$1_0 \leftarrow y_{t - 1}, $ t$1_1 \leftarrow y_t,$ t$1.used \leftarrow 2$ \\ 5163\hspace{6mm}13.5.4 $t1 \leftarrow t1 \cdot q_{i - t - 1}$ \\ 5164\hspace{6mm}13.5.5 t$2_0 \leftarrow x_{i - 2}, $ t$2_1 \leftarrow x_{i - 1}, $ t$2_2 \leftarrow x_i, $ t$2.used \leftarrow 3$ \\ 5165\hspace{6mm}13.5.6 If $\vert t1 \vert > \vert t2 \vert$ then goto step 13.5. \\ 5166\hspace{3mm}13.6 t$1 \leftarrow y \cdot q_{i - t - 1}$ \\ 5167\hspace{3mm}13.7 t$1 \leftarrow $ t$1 \cdot \beta^{i - t - 1}$ \\ 5168\hspace{3mm}13.8 $x \leftarrow x - $ t$1$ \\ 5169\hspace{3mm}13.9 If $x.sign = MP\_NEG$ then \\ 5170\hspace{6mm}13.10 t$1 \leftarrow y$ \\ 5171\hspace{6mm}13.11 t$1 \leftarrow $ t$1 \cdot \beta^{i - t - 1}$ \\ 5172\hspace{6mm}13.12 $x \leftarrow x + $ t$1$ \\ 5173\hspace{6mm}13.13 $q_{i - t - 1} \leftarrow q_{i - t - 1} - 1$ \\ 5174\\ 5175Finalize the result. \\ 517614. Clamp excess digits of $q$ \\ 517715. $c \leftarrow q, c.sign \leftarrow sign$ \\ 517816. $x.sign \leftarrow a.sign$ \\ 517917. $d \leftarrow \lfloor x / 2^{norm} \rfloor$ \\ 518018. Return(\textit{MP\_OKAY}). \\ 5181\hline 5182\end{tabular} 5183\end{center} 5184\end{small} 5185\caption{Algorithm mp\_div (continued)} 5186\end{figure} 5187\textbf{Algorithm mp\_div.} 5188This algorithm will calculate quotient and remainder from an integer division given a dividend and divisor. The algorithm is a signed 5189division and will produce a fully qualified quotient and remainder. 5190 5191First the divisor $b$ must be non-zero which is enforced in step one. If the divisor is larger than the dividend than the quotient is implicitly 5192zero and the remainder is the dividend. 5193 5194After the first two trivial cases of inputs are handled the variable $q$ is setup to receive the digits of the quotient. Two unsigned copies of the 5195divisor $y$ and dividend $x$ are made as well. The core of the division algorithm is an unsigned division and will only work if the values are 5196positive. Now the two values $x$ and $y$ must be normalized such that the leading digit of $y$ is greater than or equal to $\beta / 2$. 5197This is performed by shifting both to the left by enough bits to get the desired normalization. 5198 5199At this point the division algorithm can begin producing digits of the quotient. Recall that maximum value of the estimation used is 5200$2\beta - {2 \over \beta}$ which means that a digit of the quotient must be first produced by another means. In this case $y$ is shifted 5201to the left (\textit{step ten}) so that it has the same number of digits as $x$. The loop on step eleven will subtract multiples of the 5202shifted copy of $y$ until $x$ is smaller. Since the leading digit of $y$ is greater than or equal to $\beta/2$ this loop will iterate at most two 5203times to produce the desired leading digit of the quotient. 5204 5205Now the remainder of the digits can be produced. The equation $\hat q = \lfloor {{x_i \beta + x_{i-1}}\over y_t} \rfloor$ is used to fairly 5206accurately approximate the true quotient digit. The estimation can in theory produce an estimation as high as $2\beta - {2 \over \beta}$ but by 5207induction the upper quotient digit is correct (\textit{as established on step eleven}) and the estimate must be less than $\beta$. 5208 5209Recall from section~\ref{sec:divest} that the estimation is never too low but may be too high. The next step of the estimation process is 5210to refine the estimation. The loop on step 13.5 uses $x_i\beta^2 + x_{i-1}\beta + x_{i-2}$ and $q_{i - t - 1}(y_t\beta + y_{t-1})$ as a higher 5211order approximation to adjust the quotient digit. 5212 5213After both phases of estimation the quotient digit may still be off by a value of one\footnote{This is similar to the error introduced 5214by optimizing Barrett reduction.}. Steps 13.6 and 13.7 subtract the multiple of the divisor from the dividend (\textit{Similar to step 3.3 of 5215algorithm~\ref{fig:raddiv}} and then subsequently add a multiple of the divisor if the quotient was too large. 5216 5217Now that the quotient has been determine finializing the result is a matter of clamping the quotient, fixing the sizes and de-normalizing the 5218remainder. An important aspect of this algorithm seemingly overlooked in other descriptions such as that of Algorithm 14.20 HAC \cite[pp. 598]{HAC} 5219is that when the estimations are being made (\textit{inside the loop on step 13.5}) that the digits $y_{t-1}$, $x_{i-2}$ and $x_{i-1}$ may lie 5220outside their respective boundaries. For example, if $t = 0$ or $i \le 1$ then the digits would be undefined. In those cases the digits should 5221respectively be replaced with a zero. 5222 5223EXAM,bn_mp_div.c 5224 5225The implementation of this algorithm differs slightly from the pseudo code presented previously. In this algorithm either of the quotient $c$ or 5226remainder $d$ may be passed as a \textbf{NULL} pointer which indicates their value is not desired. For example, the C code to call the division 5227algorithm with only the quotient is 5228 5229\begin{verbatim} 5230mp_div(&a, &b, &c, NULL); /* c = [a/b] */ 5231\end{verbatim} 5232 5233Lines @108,if@ and @113,if@ handle the two trivial cases of inputs which are division by zero and dividend smaller than the divisor 5234respectively. After the two trivial cases all of the temporary variables are initialized. Line @147,neg@ determines the sign of 5235the quotient and line @148,sign@ ensures that both $x$ and $y$ are positive. 5236 5237The number of bits in the leading digit is calculated on line @151,norm@. Implictly an mp\_int with $r$ digits will require $lg(\beta)(r-1) + k$ bits 5238of precision which when reduced modulo $lg(\beta)$ produces the value of $k$. In this case $k$ is the number of bits in the leading digit which is 5239exactly what is required. For the algorithm to operate $k$ must equal $lg(\beta) - 1$ and when it does not the inputs must be normalized by shifting 5240them to the left by $lg(\beta) - 1 - k$ bits. 5241 5242Throughout the variables $n$ and $t$ will represent the highest digit of $x$ and $y$ respectively. These are first used to produce the 5243leading digit of the quotient. The loop beginning on line @184,for@ will produce the remainder of the quotient digits. 5244 5245The conditional ``continue'' on line @186,continue@ is used to prevent the algorithm from reading past the leading edge of $x$ which can occur when the 5246algorithm eliminates multiple non-zero digits in a single iteration. This ensures that $x_i$ is always non-zero since by definition the digits 5247above the $i$'th position $x$ must be zero in order for the quotient to be precise\footnote{Precise as far as integer division is concerned.}. 5248 5249Lines @214,t1@, @216,t1@ and @222,t2@ through @225,t2@ manually construct the high accuracy estimations by setting the digits of the two mp\_int 5250variables directly. 5251 5252\section{Single Digit Helpers} 5253 5254This section briefly describes a series of single digit helper algorithms which come in handy when working with small constants. All of 5255the helper functions assume the single digit input is positive and will treat them as such. 5256 5257\subsection{Single Digit Addition and Subtraction} 5258 5259Both addition and subtraction are performed by ``cheating'' and using mp\_set followed by the higher level addition or subtraction 5260algorithms. As a result these algorithms are subtantially simpler with a slight cost in performance. 5261 5262\newpage\begin{figure}[!here] 5263\begin{small} 5264\begin{center} 5265\begin{tabular}{l} 5266\hline Algorithm \textbf{mp\_add\_d}. \\ 5267\textbf{Input}. mp\_int $a$ and a mp\_digit $b$ \\ 5268\textbf{Output}. $c = a + b$ \\ 5269\hline \\ 52701. $t \leftarrow b$ (\textit{mp\_set}) \\ 52712. $c \leftarrow a + t$ \\ 52723. Return(\textit{MP\_OKAY}) \\ 5273\hline 5274\end{tabular} 5275\end{center} 5276\end{small} 5277\caption{Algorithm mp\_add\_d} 5278\end{figure} 5279 5280\textbf{Algorithm mp\_add\_d.} 5281This algorithm initiates a temporary mp\_int with the value of the single digit and uses algorithm mp\_add to add the two values together. 5282 5283EXAM,bn_mp_add_d.c 5284 5285Clever use of the letter 't'. 5286 5287\subsubsection{Subtraction} 5288The single digit subtraction algorithm mp\_sub\_d is essentially the same except it uses mp\_sub to subtract the digit from the mp\_int. 5289 5290\subsection{Single Digit Multiplication} 5291Single digit multiplication arises enough in division and radix conversion that it ought to be implement as a special case of the baseline 5292multiplication algorithm. Essentially this algorithm is a modified version of algorithm s\_mp\_mul\_digs where one of the multiplicands 5293only has one digit. 5294 5295\begin{figure}[!here] 5296\begin{small} 5297\begin{center} 5298\begin{tabular}{l} 5299\hline Algorithm \textbf{mp\_mul\_d}. \\ 5300\textbf{Input}. mp\_int $a$ and a mp\_digit $b$ \\ 5301\textbf{Output}. $c = ab$ \\ 5302\hline \\ 53031. $pa \leftarrow a.used$ \\ 53042. Grow $c$ to at least $pa + 1$ digits. \\ 53053. $oldused \leftarrow c.used$ \\ 53064. $c.used \leftarrow pa + 1$ \\ 53075. $c.sign \leftarrow a.sign$ \\ 53086. $\mu \leftarrow 0$ \\ 53097. for $ix$ from $0$ to $pa - 1$ do \\ 5310\hspace{3mm}7.1 $\hat r \leftarrow \mu + a_{ix}b$ \\ 5311\hspace{3mm}7.2 $c_{ix} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\ 5312\hspace{3mm}7.3 $\mu \leftarrow \lfloor \hat r / \beta \rfloor$ \\ 53138. $c_{pa} \leftarrow \mu$ \\ 53149. for $ix$ from $pa + 1$ to $oldused$ do \\ 5315\hspace{3mm}9.1 $c_{ix} \leftarrow 0$ \\ 531610. Clamp excess digits of $c$. \\ 531711. Return(\textit{MP\_OKAY}). \\ 5318\hline 5319\end{tabular} 5320\end{center} 5321\end{small} 5322\caption{Algorithm mp\_mul\_d} 5323\end{figure} 5324\textbf{Algorithm mp\_mul\_d.} 5325This algorithm quickly multiplies an mp\_int by a small single digit value. It is specially tailored to the job and has a minimal of overhead. 5326Unlike the full multiplication algorithms this algorithm does not require any significnat temporary storage or memory allocations. 5327 5328EXAM,bn_mp_mul_d.c 5329 5330In this implementation the destination $c$ may point to the same mp\_int as the source $a$ since the result is written after the digit is 5331read from the source. This function uses pointer aliases $tmpa$ and $tmpc$ for the digits of $a$ and $c$ respectively. 5332 5333\subsection{Single Digit Division} 5334Like the single digit multiplication algorithm, single digit division is also a fairly common algorithm used in radix conversion. Since the 5335divisor is only a single digit a specialized variant of the division algorithm can be used to compute the quotient. 5336 5337\newpage\begin{figure}[!here] 5338\begin{small} 5339\begin{center} 5340\begin{tabular}{l} 5341\hline Algorithm \textbf{mp\_div\_d}. \\ 5342\textbf{Input}. mp\_int $a$ and a mp\_digit $b$ \\ 5343\textbf{Output}. $c = \lfloor a / b \rfloor, d = a - cb$ \\ 5344\hline \\ 53451. If $b = 0$ then return(\textit{MP\_VAL}).\\ 53462. If $b = 3$ then use algorithm mp\_div\_3 instead. \\ 53473. Init $q$ to $a.used$ digits. \\ 53484. $q.used \leftarrow a.used$ \\ 53495. $q.sign \leftarrow a.sign$ \\ 53506. $\hat w \leftarrow 0$ \\ 53517. for $ix$ from $a.used - 1$ down to $0$ do \\ 5352\hspace{3mm}7.1 $\hat w \leftarrow \hat w \beta + a_{ix}$ \\ 5353\hspace{3mm}7.2 If $\hat w \ge b$ then \\ 5354\hspace{6mm}7.2.1 $t \leftarrow \lfloor \hat w / b \rfloor$ \\ 5355\hspace{6mm}7.2.2 $\hat w \leftarrow \hat w \mbox{ (mod }b\mbox{)}$ \\ 5356\hspace{3mm}7.3 else\\ 5357\hspace{6mm}7.3.1 $t \leftarrow 0$ \\ 5358\hspace{3mm}7.4 $q_{ix} \leftarrow t$ \\ 53598. $d \leftarrow \hat w$ \\ 53609. Clamp excess digits of $q$. \\ 536110. $c \leftarrow q$ \\ 536211. Return(\textit{MP\_OKAY}). \\ 5363\hline 5364\end{tabular} 5365\end{center} 5366\end{small} 5367\caption{Algorithm mp\_div\_d} 5368\end{figure} 5369\textbf{Algorithm mp\_div\_d.} 5370This algorithm divides the mp\_int $a$ by the single mp\_digit $b$ using an optimized approach. Essentially in every iteration of the 5371algorithm another digit of the dividend is reduced and another digit of quotient produced. Provided $b < \beta$ the value of $\hat w$ 5372after step 7.1 will be limited such that $0 \le \lfloor \hat w / b \rfloor < \beta$. 5373 5374If the divisor $b$ is equal to three a variant of this algorithm is used which is called mp\_div\_3. It replaces the division by three with 5375a multiplication by $\lfloor \beta / 3 \rfloor$ and the appropriate shift and residual fixup. In essence it is much like the Barrett reduction 5376from chapter seven. 5377 5378EXAM,bn_mp_div_d.c 5379 5380Like the implementation of algorithm mp\_div this algorithm allows either of the quotient or remainder to be passed as a \textbf{NULL} pointer to 5381indicate the respective value is not required. This allows a trivial single digit modular reduction algorithm, mp\_mod\_d to be created. 5382 5383The division and remainder on lines @44,/@ and @45,%@ can be replaced often by a single division on most processors. For example, the 32-bit x86 based 5384processors can divide a 64-bit quantity by a 32-bit quantity and produce the quotient and remainder simultaneously. Unfortunately the GCC 5385compiler does not recognize that optimization and will actually produce two function calls to find the quotient and remainder respectively. 5386 5387\subsection{Single Digit Root Extraction} 5388 5389Finding the $n$'th root of an integer is fairly easy as far as numerical analysis is concerned. Algorithms such as the Newton-Raphson approximation 5390(\ref{eqn:newton}) series will converge very quickly to a root for any continuous function $f(x)$. 5391 5392\begin{equation} 5393x_{i+1} = x_i - {f(x_i) \over f'(x_i)} 5394\label{eqn:newton} 5395\end{equation} 5396 5397In this case the $n$'th root is desired and $f(x) = x^n - a$ where $a$ is the integer of which the root is desired. The derivative of $f(x)$ is 5398simply $f'(x) = nx^{n - 1}$. Of particular importance is that this algorithm will be used over the integers not over the a more continuous domain 5399such as the real numbers. As a result the root found can be above the true root by few and must be manually adjusted. Ideally at the end of the 5400algorithm the $n$'th root $b$ of an integer $a$ is desired such that $b^n \le a$. 5401 5402\newpage\begin{figure}[!here] 5403\begin{small} 5404\begin{center} 5405\begin{tabular}{l} 5406\hline Algorithm \textbf{mp\_n\_root}. \\ 5407\textbf{Input}. mp\_int $a$ and a mp\_digit $b$ \\ 5408\textbf{Output}. $c^b \le a$ \\ 5409\hline \\ 54101. If $b$ is even and $a.sign = MP\_NEG$ return(\textit{MP\_VAL}). \\ 54112. $sign \leftarrow a.sign$ \\ 54123. $a.sign \leftarrow MP\_ZPOS$ \\ 54134. t$2 \leftarrow 2$ \\ 54145. Loop \\ 5415\hspace{3mm}5.1 t$1 \leftarrow $ t$2$ \\ 5416\hspace{3mm}5.2 t$3 \leftarrow $ t$1^{b - 1}$ \\ 5417\hspace{3mm}5.3 t$2 \leftarrow $ t$3 $ $\cdot$ t$1$ \\ 5418\hspace{3mm}5.4 t$2 \leftarrow $ t$2 - a$ \\ 5419\hspace{3mm}5.5 t$3 \leftarrow $ t$3 \cdot b$ \\ 5420\hspace{3mm}5.6 t$3 \leftarrow \lfloor $t$2 / $t$3 \rfloor$ \\ 5421\hspace{3mm}5.7 t$2 \leftarrow $ t$1 - $ t$3$ \\ 5422\hspace{3mm}5.8 If t$1 \ne $ t$2$ then goto step 5. \\ 54236. Loop \\ 5424\hspace{3mm}6.1 t$2 \leftarrow $ t$1^b$ \\ 5425\hspace{3mm}6.2 If t$2 > a$ then \\ 5426\hspace{6mm}6.2.1 t$1 \leftarrow $ t$1 - 1$ \\ 5427\hspace{6mm}6.2.2 Goto step 6. \\ 54287. $a.sign \leftarrow sign$ \\ 54298. $c \leftarrow $ t$1$ \\ 54309. $c.sign \leftarrow sign$ \\ 543110. Return(\textit{MP\_OKAY}). \\ 5432\hline 5433\end{tabular} 5434\end{center} 5435\end{small} 5436\caption{Algorithm mp\_n\_root} 5437\end{figure} 5438\textbf{Algorithm mp\_n\_root.} 5439This algorithm finds the integer $n$'th root of an input using the Newton-Raphson approach. It is partially optimized based on the observation 5440that the numerator of ${f(x) \over f'(x)}$ can be derived from a partial denominator. That is at first the denominator is calculated by finding 5441$x^{b - 1}$. This value can then be multiplied by $x$ and have $a$ subtracted from it to find the numerator. This saves a total of $b - 1$ 5442multiplications by t$1$ inside the loop. 5443 5444The initial value of the approximation is t$2 = 2$ which allows the algorithm to start with very small values and quickly converge on the 5445root. Ideally this algorithm is meant to find the $n$'th root of an input where $n$ is bounded by $2 \le n \le 5$. 5446 5447EXAM,bn_mp_n_root.c 5448 5449\section{Random Number Generation} 5450 5451Random numbers come up in a variety of activities from public key cryptography to simple simulations and various randomized algorithms. Pollard-Rho 5452factoring for example, can make use of random values as starting points to find factors of a composite integer. In this case the algorithm presented 5453is solely for simulations and not intended for cryptographic use. 5454 5455\newpage\begin{figure}[!here] 5456\begin{small} 5457\begin{center} 5458\begin{tabular}{l} 5459\hline Algorithm \textbf{mp\_rand}. \\ 5460\textbf{Input}. An integer $b$ \\ 5461\textbf{Output}. A pseudo-random number of $b$ digits \\ 5462\hline \\ 54631. $a \leftarrow 0$ \\ 54642. If $b \le 0$ return(\textit{MP\_OKAY}) \\ 54653. Pick a non-zero random digit $d$. \\ 54664. $a \leftarrow a + d$ \\ 54675. for $ix$ from 1 to $d - 1$ do \\ 5468\hspace{3mm}5.1 $a \leftarrow a \cdot \beta$ \\ 5469\hspace{3mm}5.2 Pick a random digit $d$. \\ 5470\hspace{3mm}5.3 $a \leftarrow a + d$ \\ 54716. Return(\textit{MP\_OKAY}). \\ 5472\hline 5473\end{tabular} 5474\end{center} 5475\end{small} 5476\caption{Algorithm mp\_rand} 5477\end{figure} 5478\textbf{Algorithm mp\_rand.} 5479This algorithm produces a pseudo-random integer of $b$ digits. By ensuring that the first digit is non-zero the algorithm also guarantees that the 5480final result has at least $b$ digits. It relies heavily on a third-part random number generator which should ideally generate uniformly all of 5481the integers from $0$ to $\beta - 1$. 5482 5483EXAM,bn_mp_rand.c 5484 5485\section{Formatted Representations} 5486The ability to emit a radix-$n$ textual representation of an integer is useful for interacting with human parties. For example, the ability to 5487be given a string of characters such as ``114585'' and turn it into the radix-$\beta$ equivalent would make it easier to enter numbers 5488into a program. 5489 5490\subsection{Reading Radix-n Input} 5491For the purposes of this text we will assume that a simple lower ASCII map (\ref{fig:ASC}) is used for the values of from $0$ to $63$ to 5492printable characters. For example, when the character ``N'' is read it represents the integer $23$. The first $16$ characters of the 5493map are for the common representations up to hexadecimal. After that they match the ``base64'' encoding scheme which are suitable chosen 5494such that they are printable. While outputting as base64 may not be too helpful for human operators it does allow communication via non binary 5495mediums. 5496 5497\newpage\begin{figure}[here] 5498\begin{center} 5499\begin{tabular}{cc|cc|cc|cc} 5500\hline \textbf{Value} & \textbf{Char} & \textbf{Value} & \textbf{Char} & \textbf{Value} & \textbf{Char} & \textbf{Value} & \textbf{Char} \\ 5501\hline 55020 & 0 & 1 & 1 & 2 & 2 & 3 & 3 \\ 55034 & 4 & 5 & 5 & 6 & 6 & 7 & 7 \\ 55048 & 8 & 9 & 9 & 10 & A & 11 & B \\ 550512 & C & 13 & D & 14 & E & 15 & F \\ 550616 & G & 17 & H & 18 & I & 19 & J \\ 550720 & K & 21 & L & 22 & M & 23 & N \\ 550824 & O & 25 & P & 26 & Q & 27 & R \\ 550928 & S & 29 & T & 30 & U & 31 & V \\ 551032 & W & 33 & X & 34 & Y & 35 & Z \\ 551136 & a & 37 & b & 38 & c & 39 & d \\ 551240 & e & 41 & f & 42 & g & 43 & h \\ 551344 & i & 45 & j & 46 & k & 47 & l \\ 551448 & m & 49 & n & 50 & o & 51 & p \\ 551552 & q & 53 & r & 54 & s & 55 & t \\ 551656 & u & 57 & v & 58 & w & 59 & x \\ 551760 & y & 61 & z & 62 & $+$ & 63 & $/$ \\ 5518\hline 5519\end{tabular} 5520\end{center} 5521\caption{Lower ASCII Map} 5522\label{fig:ASC} 5523\end{figure} 5524 5525\newpage\begin{figure}[!here] 5526\begin{small} 5527\begin{center} 5528\begin{tabular}{l} 5529\hline Algorithm \textbf{mp\_read\_radix}. \\ 5530\textbf{Input}. A string $str$ of length $sn$ and radix $r$. \\ 5531\textbf{Output}. The radix-$\beta$ equivalent mp\_int. \\ 5532\hline \\ 55331. If $r < 2$ or $r > 64$ return(\textit{MP\_VAL}). \\ 55342. $ix \leftarrow 0$ \\ 55353. If $str_0 =$ ``-'' then do \\ 5536\hspace{3mm}3.1 $ix \leftarrow ix + 1$ \\ 5537\hspace{3mm}3.2 $sign \leftarrow MP\_NEG$ \\ 55384. else \\ 5539\hspace{3mm}4.1 $sign \leftarrow MP\_ZPOS$ \\ 55405. $a \leftarrow 0$ \\ 55416. for $iy$ from $ix$ to $sn - 1$ do \\ 5542\hspace{3mm}6.1 Let $y$ denote the position in the map of $str_{iy}$. \\ 5543\hspace{3mm}6.2 If $str_{iy}$ is not in the map or $y \ge r$ then goto step 7. \\ 5544\hspace{3mm}6.3 $a \leftarrow a \cdot r$ \\ 5545\hspace{3mm}6.4 $a \leftarrow a + y$ \\ 55467. If $a \ne 0$ then $a.sign \leftarrow sign$ \\ 55478. Return(\textit{MP\_OKAY}). \\ 5548\hline 5549\end{tabular} 5550\end{center} 5551\end{small} 5552\caption{Algorithm mp\_read\_radix} 5553\end{figure} 5554\textbf{Algorithm mp\_read\_radix.} 5555This algorithm will read an ASCII string and produce the radix-$\beta$ mp\_int representation of the same integer. A minus symbol ``-'' may precede the 5556string to indicate the value is negative, otherwise it is assumed to be positive. The algorithm will read up to $sn$ characters from the input 5557and will stop when it reads a character it cannot map the algorithm stops reading characters from the string. This allows numbers to be embedded 5558as part of larger input without any significant problem. 5559 5560EXAM,bn_mp_read_radix.c 5561 5562\subsection{Generating Radix-$n$ Output} 5563Generating radix-$n$ output is fairly trivial with a division and remainder algorithm. 5564 5565\newpage\begin{figure}[!here] 5566\begin{small} 5567\begin{center} 5568\begin{tabular}{l} 5569\hline Algorithm \textbf{mp\_toradix}. \\ 5570\textbf{Input}. A mp\_int $a$ and an integer $r$\\ 5571\textbf{Output}. The radix-$r$ representation of $a$ \\ 5572\hline \\ 55731. If $r < 2$ or $r > 64$ return(\textit{MP\_VAL}). \\ 55742. If $a = 0$ then $str = $ ``$0$'' and return(\textit{MP\_OKAY}). \\ 55753. $t \leftarrow a$ \\ 55764. $str \leftarrow$ ``'' \\ 55775. if $t.sign = MP\_NEG$ then \\ 5578\hspace{3mm}5.1 $str \leftarrow str + $ ``-'' \\ 5579\hspace{3mm}5.2 $t.sign = MP\_ZPOS$ \\ 55806. While ($t \ne 0$) do \\ 5581\hspace{3mm}6.1 $d \leftarrow t \mbox{ (mod }r\mbox{)}$ \\ 5582\hspace{3mm}6.2 $t \leftarrow \lfloor t / r \rfloor$ \\ 5583\hspace{3mm}6.3 Look up $d$ in the map and store the equivalent character in $y$. \\ 5584\hspace{3mm}6.4 $str \leftarrow str + y$ \\ 55857. If $str_0 = $``$-$'' then \\ 5586\hspace{3mm}7.1 Reverse the digits $str_1, str_2, \ldots str_n$. \\ 55878. Otherwise \\ 5588\hspace{3mm}8.1 Reverse the digits $str_0, str_1, \ldots str_n$. \\ 55899. Return(\textit{MP\_OKAY}).\\ 5590\hline 5591\end{tabular} 5592\end{center} 5593\end{small} 5594\caption{Algorithm mp\_toradix} 5595\end{figure} 5596\textbf{Algorithm mp\_toradix.} 5597This algorithm computes the radix-$r$ representation of an mp\_int $a$. The ``digits'' of the representation are extracted by reducing 5598successive powers of $\lfloor a / r^k \rfloor$ the input modulo $r$ until $r^k > a$. Note that instead of actually dividing by $r^k$ in 5599each iteration the quotient $\lfloor a / r \rfloor$ is saved for the next iteration. As a result a series of trivial $n \times 1$ divisions 5600are required instead of a series of $n \times k$ divisions. One design flaw of this approach is that the digits are produced in the reverse order 5601(see~\ref{fig:mpradix}). To remedy this flaw the digits must be swapped or simply ``reversed''. 5602 5603\begin{figure} 5604\begin{center} 5605\begin{tabular}{|c|c|c|} 5606\hline \textbf{Value of $a$} & \textbf{Value of $d$} & \textbf{Value of $str$} \\ 5607\hline $1234$ & -- & -- \\ 5608\hline $123$ & $4$ & ``4'' \\ 5609\hline $12$ & $3$ & ``43'' \\ 5610\hline $1$ & $2$ & ``432'' \\ 5611\hline $0$ & $1$ & ``4321'' \\ 5612\hline 5613\end{tabular} 5614\end{center} 5615\caption{Example of Algorithm mp\_toradix.} 5616\label{fig:mpradix} 5617\end{figure} 5618 5619EXAM,bn_mp_toradix.c 5620 5621\chapter{Number Theoretic Algorithms} 5622This chapter discusses several fundamental number theoretic algorithms such as the greatest common divisor, least common multiple and Jacobi 5623symbol computation. These algorithms arise as essential components in several key cryptographic algorithms such as the RSA public key algorithm and 5624various Sieve based factoring algorithms. 5625 5626\section{Greatest Common Divisor} 5627The greatest common divisor of two integers $a$ and $b$, often denoted as $(a, b)$ is the largest integer $k$ that is a proper divisor of 5628both $a$ and $b$. That is, $k$ is the largest integer such that $0 \equiv a \mbox{ (mod }k\mbox{)}$ and $0 \equiv b \mbox{ (mod }k\mbox{)}$ occur 5629simultaneously. 5630 5631The most common approach (cite) is to reduce one input modulo another. That is if $a$ and $b$ are divisible by some integer $k$ and if $qa + r = b$ then 5632$r$ is also divisible by $k$. The reduction pattern follows $\left < a , b \right > \rightarrow \left < b, a \mbox{ mod } b \right >$. 5633 5634\newpage\begin{figure}[!here] 5635\begin{small} 5636\begin{center} 5637\begin{tabular}{l} 5638\hline Algorithm \textbf{Greatest Common Divisor (I)}. \\ 5639\textbf{Input}. Two positive integers $a$ and $b$ greater than zero. \\ 5640\textbf{Output}. The greatest common divisor $(a, b)$. \\ 5641\hline \\ 56421. While ($b > 0$) do \\ 5643\hspace{3mm}1.1 $r \leftarrow a \mbox{ (mod }b\mbox{)}$ \\ 5644\hspace{3mm}1.2 $a \leftarrow b$ \\ 5645\hspace{3mm}1.3 $b \leftarrow r$ \\ 56462. Return($a$). \\ 5647\hline 5648\end{tabular} 5649\end{center} 5650\end{small} 5651\caption{Algorithm Greatest Common Divisor (I)} 5652\label{fig:gcd1} 5653\end{figure} 5654 5655This algorithm will quickly converge on the greatest common divisor since the residue $r$ tends diminish rapidly. However, divisions are 5656relatively expensive operations to perform and should ideally be avoided. There is another approach based on a similar relationship of 5657greatest common divisors. The faster approach is based on the observation that if $k$ divides both $a$ and $b$ it will also divide $a - b$. 5658In particular, we would like $a - b$ to decrease in magnitude which implies that $b \ge a$. 5659 5660\begin{figure}[!here] 5661\begin{small} 5662\begin{center} 5663\begin{tabular}{l} 5664\hline Algorithm \textbf{Greatest Common Divisor (II)}. \\ 5665\textbf{Input}. Two positive integers $a$ and $b$ greater than zero. \\ 5666\textbf{Output}. The greatest common divisor $(a, b)$. \\ 5667\hline \\ 56681. While ($b > 0$) do \\ 5669\hspace{3mm}1.1 Swap $a$ and $b$ such that $a$ is the smallest of the two. \\ 5670\hspace{3mm}1.2 $b \leftarrow b - a$ \\ 56712. Return($a$). \\ 5672\hline 5673\end{tabular} 5674\end{center} 5675\end{small} 5676\caption{Algorithm Greatest Common Divisor (II)} 5677\label{fig:gcd2} 5678\end{figure} 5679 5680\textbf{Proof} \textit{Algorithm~\ref{fig:gcd2} will return the greatest common divisor of $a$ and $b$.} 5681The algorithm in figure~\ref{fig:gcd2} will eventually terminate since $b \ge a$ the subtraction in step 1.2 will be a value less than $b$. In other 5682words in every iteration that tuple $\left < a, b \right >$ decrease in magnitude until eventually $a = b$. Since both $a$ and $b$ are always 5683divisible by the greatest common divisor (\textit{until the last iteration}) and in the last iteration of the algorithm $b = 0$, therefore, in the 5684second to last iteration of the algorithm $b = a$ and clearly $(a, a) = a$ which concludes the proof. \textbf{QED}. 5685 5686As a matter of practicality algorithm \ref{fig:gcd1} decreases far too slowly to be useful. Specially if $b$ is much larger than $a$ such that 5687$b - a$ is still very much larger than $a$. A simple addition to the algorithm is to divide $b - a$ by a power of some integer $p$ which does 5688not divide the greatest common divisor but will divide $b - a$. In this case ${b - a} \over p$ is also an integer and still divisible by 5689the greatest common divisor. 5690 5691However, instead of factoring $b - a$ to find a suitable value of $p$ the powers of $p$ can be removed from $a$ and $b$ that are in common first. 5692Then inside the loop whenever $b - a$ is divisible by some power of $p$ it can be safely removed. 5693 5694\begin{figure}[!here] 5695\begin{small} 5696\begin{center} 5697\begin{tabular}{l} 5698\hline Algorithm \textbf{Greatest Common Divisor (III)}. \\ 5699\textbf{Input}. Two positive integers $a$ and $b$ greater than zero. \\ 5700\textbf{Output}. The greatest common divisor $(a, b)$. \\ 5701\hline \\ 57021. $k \leftarrow 0$ \\ 57032. While $a$ and $b$ are both divisible by $p$ do \\ 5704\hspace{3mm}2.1 $a \leftarrow \lfloor a / p \rfloor$ \\ 5705\hspace{3mm}2.2 $b \leftarrow \lfloor b / p \rfloor$ \\ 5706\hspace{3mm}2.3 $k \leftarrow k + 1$ \\ 57073. While $a$ is divisible by $p$ do \\ 5708\hspace{3mm}3.1 $a \leftarrow \lfloor a / p \rfloor$ \\ 57094. While $b$ is divisible by $p$ do \\ 5710\hspace{3mm}4.1 $b \leftarrow \lfloor b / p \rfloor$ \\ 57115. While ($b > 0$) do \\ 5712\hspace{3mm}5.1 Swap $a$ and $b$ such that $a$ is the smallest of the two. \\ 5713\hspace{3mm}5.2 $b \leftarrow b - a$ \\ 5714\hspace{3mm}5.3 While $b$ is divisible by $p$ do \\ 5715\hspace{6mm}5.3.1 $b \leftarrow \lfloor b / p \rfloor$ \\ 57166. Return($a \cdot p^k$). \\ 5717\hline 5718\end{tabular} 5719\end{center} 5720\end{small} 5721\caption{Algorithm Greatest Common Divisor (III)} 5722\label{fig:gcd3} 5723\end{figure} 5724 5725This algorithm is based on the first except it removes powers of $p$ first and inside the main loop to ensure the tuple $\left < a, b \right >$ 5726decreases more rapidly. The first loop on step two removes powers of $p$ that are in common. A count, $k$, is kept which will present a common 5727divisor of $p^k$. After step two the remaining common divisor of $a$ and $b$ cannot be divisible by $p$. This means that $p$ can be safely 5728divided out of the difference $b - a$ so long as the division leaves no remainder. 5729 5730In particular the value of $p$ should be chosen such that the division on step 5.3.1 occur often. It also helps that division by $p$ be easy 5731to compute. The ideal choice of $p$ is two since division by two amounts to a right logical shift. Another important observation is that by 5732step five both $a$ and $b$ are odd. Therefore, the diffrence $b - a$ must be even which means that each iteration removes one bit from the 5733largest of the pair. 5734 5735\subsection{Complete Greatest Common Divisor} 5736The algorithms presented so far cannot handle inputs which are zero or negative. The following algorithm can handle all input cases properly 5737and will produce the greatest common divisor. 5738 5739\newpage\begin{figure}[!here] 5740\begin{small} 5741\begin{center} 5742\begin{tabular}{l} 5743\hline Algorithm \textbf{mp\_gcd}. \\ 5744\textbf{Input}. mp\_int $a$ and $b$ \\ 5745\textbf{Output}. The greatest common divisor $c = (a, b)$. \\ 5746\hline \\ 57471. If $a = 0$ then \\ 5748\hspace{3mm}1.1 $c \leftarrow \vert b \vert $ \\ 5749\hspace{3mm}1.2 Return(\textit{MP\_OKAY}). \\ 57502. If $b = 0$ then \\ 5751\hspace{3mm}2.1 $c \leftarrow \vert a \vert $ \\ 5752\hspace{3mm}2.2 Return(\textit{MP\_OKAY}). \\ 57533. $u \leftarrow \vert a \vert, v \leftarrow \vert b \vert$ \\ 57544. $k \leftarrow 0$ \\ 57555. While $u.used > 0$ and $v.used > 0$ and $u_0 \equiv v_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\ 5756\hspace{3mm}5.1 $k \leftarrow k + 1$ \\ 5757\hspace{3mm}5.2 $u \leftarrow \lfloor u / 2 \rfloor$ \\ 5758\hspace{3mm}5.3 $v \leftarrow \lfloor v / 2 \rfloor$ \\ 57596. While $u.used > 0$ and $u_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\ 5760\hspace{3mm}6.1 $u \leftarrow \lfloor u / 2 \rfloor$ \\ 57617. While $v.used > 0$ and $v_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\ 5762\hspace{3mm}7.1 $v \leftarrow \lfloor v / 2 \rfloor$ \\ 57638. While $v.used > 0$ \\ 5764\hspace{3mm}8.1 If $\vert u \vert > \vert v \vert$ then \\ 5765\hspace{6mm}8.1.1 Swap $u$ and $v$. \\ 5766\hspace{3mm}8.2 $v \leftarrow \vert v \vert - \vert u \vert$ \\ 5767\hspace{3mm}8.3 While $v.used > 0$ and $v_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\ 5768\hspace{6mm}8.3.1 $v \leftarrow \lfloor v / 2 \rfloor$ \\ 57699. $c \leftarrow u \cdot 2^k$ \\ 577010. Return(\textit{MP\_OKAY}). \\ 5771\hline 5772\end{tabular} 5773\end{center} 5774\end{small} 5775\caption{Algorithm mp\_gcd} 5776\end{figure} 5777\textbf{Algorithm mp\_gcd.} 5778This algorithm will produce the greatest common divisor of two mp\_ints $a$ and $b$. The algorithm was originally based on Algorithm B of 5779Knuth \cite[pp. 338]{TAOCPV2} but has been modified to be simpler to explain. In theory it achieves the same asymptotic working time as 5780Algorithm B and in practice this appears to be true. 5781 5782The first two steps handle the cases where either one of or both inputs are zero. If either input is zero the greatest common divisor is the 5783largest input or zero if they are both zero. If the inputs are not trivial than $u$ and $v$ are assigned the absolute values of 5784$a$ and $b$ respectively and the algorithm will proceed to reduce the pair. 5785 5786Step five will divide out any common factors of two and keep track of the count in the variable $k$. After this step, two is no longer a 5787factor of the remaining greatest common divisor between $u$ and $v$ and can be safely evenly divided out of either whenever they are even. Step 5788six and seven ensure that the $u$ and $v$ respectively have no more factors of two. At most only one of the while--loops will iterate since 5789they cannot both be even. 5790 5791By step eight both of $u$ and $v$ are odd which is required for the inner logic. First the pair are swapped such that $v$ is equal to 5792or greater than $u$. This ensures that the subtraction on step 8.2 will always produce a positive and even result. Step 8.3 removes any 5793factors of two from the difference $u$ to ensure that in the next iteration of the loop both are once again odd. 5794 5795After $v = 0$ occurs the variable $u$ has the greatest common divisor of the pair $\left < u, v \right >$ just after step six. The result 5796must be adjusted by multiplying by the common factors of two ($2^k$) removed earlier. 5797 5798EXAM,bn_mp_gcd.c 5799 5800This function makes use of the macros mp\_iszero and mp\_iseven. The former evaluates to $1$ if the input mp\_int is equivalent to the 5801integer zero otherwise it evaluates to $0$. The latter evaluates to $1$ if the input mp\_int represents a non-zero even integer otherwise 5802it evaluates to $0$. Note that just because mp\_iseven may evaluate to $0$ does not mean the input is odd, it could also be zero. The three 5803trivial cases of inputs are handled on lines @23,zero@ through @29,}@. After those lines the inputs are assumed to be non-zero. 5804 5805Lines @32,if@ and @36,if@ make local copies $u$ and $v$ of the inputs $a$ and $b$ respectively. At this point the common factors of two 5806must be divided out of the two inputs. The block starting at line @43,common@ removes common factors of two by first counting the number of trailing 5807zero bits in both. The local integer $k$ is used to keep track of how many factors of $2$ are pulled out of both values. It is assumed that 5808the number of factors will not exceed the maximum value of a C ``int'' data type\footnote{Strictly speaking no array in C may have more than 5809entries than are accessible by an ``int'' so this is not a limitation.}. 5810 5811At this point there are no more common factors of two in the two values. The divisions by a power of two on lines @60,div_2d@ and @67,div_2d@ remove 5812any independent factors of two such that both $u$ and $v$ are guaranteed to be an odd integer before hitting the main body of the algorithm. The while loop 5813on line @72, while@ performs the reduction of the pair until $v$ is equal to zero. The unsigned comparison and subtraction algorithms are used in 5814place of the full signed routines since both values are guaranteed to be positive and the result of the subtraction is guaranteed to be non-negative. 5815 5816\section{Least Common Multiple} 5817The least common multiple of a pair of integers is their product divided by their greatest common divisor. For two integers $a$ and $b$ the 5818least common multiple is normally denoted as $[ a, b ]$ and numerically equivalent to ${ab} \over {(a, b)}$. For example, if $a = 2 \cdot 2 \cdot 3 = 12$ 5819and $b = 2 \cdot 3 \cdot 3 \cdot 7 = 126$ the least common multiple is ${126 \over {(12, 126)}} = {126 \over 6} = 21$. 5820 5821The least common multiple arises often in coding theory as well as number theory. If two functions have periods of $a$ and $b$ respectively they will 5822collide, that is be in synchronous states, after only $[ a, b ]$ iterations. This is why, for example, random number generators based on 5823Linear Feedback Shift Registers (LFSR) tend to use registers with periods which are co-prime (\textit{e.g. the greatest common divisor is one.}). 5824Similarly in number theory if a composite $n$ has two prime factors $p$ and $q$ then maximal order of any unit of $\Z/n\Z$ will be $[ p - 1, q - 1] $. 5825 5826\begin{figure}[!here] 5827\begin{small} 5828\begin{center} 5829\begin{tabular}{l} 5830\hline Algorithm \textbf{mp\_lcm}. \\ 5831\textbf{Input}. mp\_int $a$ and $b$ \\ 5832\textbf{Output}. The least common multiple $c = [a, b]$. \\ 5833\hline \\ 58341. $c \leftarrow (a, b)$ \\ 58352. $t \leftarrow a \cdot b$ \\ 58363. $c \leftarrow \lfloor t / c \rfloor$ \\ 58374. Return(\textit{MP\_OKAY}). \\ 5838\hline 5839\end{tabular} 5840\end{center} 5841\end{small} 5842\caption{Algorithm mp\_lcm} 5843\end{figure} 5844\textbf{Algorithm mp\_lcm.} 5845This algorithm computes the least common multiple of two mp\_int inputs $a$ and $b$. It computes the least common multiple directly by 5846dividing the product of the two inputs by their greatest common divisor. 5847 5848EXAM,bn_mp_lcm.c 5849 5850\section{Jacobi Symbol Computation} 5851To explain the Jacobi Symbol we shall first discuss the Legendre function\footnote{Arrg. What is the name of this?} off which the Jacobi symbol is 5852defined. The Legendre function computes whether or not an integer $a$ is a quadratic residue modulo an odd prime $p$. Numerically it is 5853equivalent to equation \ref{eqn:legendre}. 5854 5855\textit{-- Tom, don't be an ass, cite your source here...!} 5856 5857\begin{equation} 5858a^{(p-1)/2} \equiv \begin{array}{rl} 5859 -1 & \mbox{if }a\mbox{ is a quadratic non-residue.} \\ 5860 0 & \mbox{if }a\mbox{ divides }p\mbox{.} \\ 5861 1 & \mbox{if }a\mbox{ is a quadratic residue}. 5862 \end{array} \mbox{ (mod }p\mbox{)} 5863\label{eqn:legendre} 5864\end{equation} 5865 5866\textbf{Proof.} \textit{Equation \ref{eqn:legendre} correctly identifies the residue status of an integer $a$ modulo a prime $p$.} 5867An integer $a$ is a quadratic residue if the following equation has a solution. 5868 5869\begin{equation} 5870x^2 \equiv a \mbox{ (mod }p\mbox{)} 5871\label{eqn:root} 5872\end{equation} 5873 5874Consider the following equation. 5875 5876\begin{equation} 58770 \equiv x^{p-1} - 1 \equiv \left \lbrace \left (x^2 \right )^{(p-1)/2} - a^{(p-1)/2} \right \rbrace + \left ( a^{(p-1)/2} - 1 \right ) \mbox{ (mod }p\mbox{)} 5878\label{eqn:rooti} 5879\end{equation} 5880 5881Whether equation \ref{eqn:root} has a solution or not equation \ref{eqn:rooti} is always true. If $a^{(p-1)/2} - 1 \equiv 0 \mbox{ (mod }p\mbox{)}$ 5882then the quantity in the braces must be zero. By reduction, 5883 5884\begin{eqnarray} 5885\left (x^2 \right )^{(p-1)/2} - a^{(p-1)/2} \equiv 0 \nonumber \\ 5886\left (x^2 \right )^{(p-1)/2} \equiv a^{(p-1)/2} \nonumber \\ 5887x^2 \equiv a \mbox{ (mod }p\mbox{)} 5888\end{eqnarray} 5889 5890As a result there must be a solution to the quadratic equation and in turn $a$ must be a quadratic residue. If $a$ does not divide $p$ and $a$ 5891is not a quadratic residue then the only other value $a^{(p-1)/2}$ may be congruent to is $-1$ since 5892\begin{equation} 58930 \equiv a^{p - 1} - 1 \equiv (a^{(p-1)/2} + 1)(a^{(p-1)/2} - 1) \mbox{ (mod }p\mbox{)} 5894\end{equation} 5895One of the terms on the right hand side must be zero. \textbf{QED} 5896 5897\subsection{Jacobi Symbol} 5898The Jacobi symbol is a generalization of the Legendre function for any odd non prime moduli $p$ greater than 2. If $p = \prod_{i=0}^n p_i$ then 5899the Jacobi symbol $\left ( { a \over p } \right )$ is equal to the following equation. 5900 5901\begin{equation} 5902\left ( { a \over p } \right ) = \left ( { a \over p_0} \right ) \left ( { a \over p_1} \right ) \ldots \left ( { a \over p_n} \right ) 5903\end{equation} 5904 5905By inspection if $p$ is prime the Jacobi symbol is equivalent to the Legendre function. The following facts\footnote{See HAC \cite[pp. 72-74]{HAC} for 5906further details.} will be used to derive an efficient Jacobi symbol algorithm. Where $p$ is an odd integer greater than two and $a, b \in \Z$ the 5907following are true. 5908 5909\begin{enumerate} 5910\item $\left ( { a \over p} \right )$ equals $-1$, $0$ or $1$. 5911\item $\left ( { ab \over p} \right ) = \left ( { a \over p} \right )\left ( { b \over p} \right )$. 5912\item If $a \equiv b$ then $\left ( { a \over p} \right ) = \left ( { b \over p} \right )$. 5913\item $\left ( { 2 \over p} \right )$ equals $1$ if $p \equiv 1$ or $7 \mbox{ (mod }8\mbox{)}$. Otherwise, it equals $-1$. 5914\item $\left ( { a \over p} \right ) \equiv \left ( { p \over a} \right ) \cdot (-1)^{(p-1)(a-1)/4}$. More specifically 5915$\left ( { a \over p} \right ) = \left ( { p \over a} \right )$ if $p \equiv a \equiv 1 \mbox{ (mod }4\mbox{)}$. 5916\end{enumerate} 5917 5918Using these facts if $a = 2^k \cdot a'$ then 5919 5920\begin{eqnarray} 5921\left ( { a \over p } \right ) = \left ( {{2^k} \over p } \right ) \left ( {a' \over p} \right ) \nonumber \\ 5922 = \left ( {2 \over p } \right )^k \left ( {a' \over p} \right ) 5923\label{eqn:jacobi} 5924\end{eqnarray} 5925 5926By fact five, 5927 5928\begin{equation} 5929\left ( { a \over p } \right ) = \left ( { p \over a } \right ) \cdot (-1)^{(p-1)(a-1)/4} 5930\end{equation} 5931 5932Subsequently by fact three since $p \equiv (p \mbox{ mod }a) \mbox{ (mod }a\mbox{)}$ then 5933 5934\begin{equation} 5935\left ( { a \over p } \right ) = \left ( { {p \mbox{ mod } a} \over a } \right ) \cdot (-1)^{(p-1)(a-1)/4} 5936\end{equation} 5937 5938By putting both observations into equation \ref{eqn:jacobi} the following simplified equation is formed. 5939 5940\begin{equation} 5941\left ( { a \over p } \right ) = \left ( {2 \over p } \right )^k \left ( {{p\mbox{ mod }a'} \over a'} \right ) \cdot (-1)^{(p-1)(a'-1)/4} 5942\end{equation} 5943 5944The value of $\left ( {{p \mbox{ mod }a'} \over a'} \right )$ can be found by using the same equation recursively. The value of 5945$\left ( {2 \over p } \right )^k$ equals $1$ if $k$ is even otherwise it equals $\left ( {2 \over p } \right )$. Using this approach the 5946factors of $p$ do not have to be known. Furthermore, if $(a, p) = 1$ then the algorithm will terminate when the recursion requests the 5947Jacobi symbol computation of $\left ( {1 \over a'} \right )$ which is simply $1$. 5948 5949\newpage\begin{figure}[!here] 5950\begin{small} 5951\begin{center} 5952\begin{tabular}{l} 5953\hline Algorithm \textbf{mp\_jacobi}. \\ 5954\textbf{Input}. mp\_int $a$ and $p$, $a \ge 0$, $p \ge 3$, $p \equiv 1 \mbox{ (mod }2\mbox{)}$ \\ 5955\textbf{Output}. The Jacobi symbol $c = \left ( {a \over p } \right )$. \\ 5956\hline \\ 59571. If $a = 0$ then \\ 5958\hspace{3mm}1.1 $c \leftarrow 0$ \\ 5959\hspace{3mm}1.2 Return(\textit{MP\_OKAY}). \\ 59602. If $a = 1$ then \\ 5961\hspace{3mm}2.1 $c \leftarrow 1$ \\ 5962\hspace{3mm}2.2 Return(\textit{MP\_OKAY}). \\ 59633. $a' \leftarrow a$ \\ 59644. $k \leftarrow 0$ \\ 59655. While $a'.used > 0$ and $a'_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\ 5966\hspace{3mm}5.1 $k \leftarrow k + 1$ \\ 5967\hspace{3mm}5.2 $a' \leftarrow \lfloor a' / 2 \rfloor$ \\ 59686. If $k \equiv 0 \mbox{ (mod }2\mbox{)}$ then \\ 5969\hspace{3mm}6.1 $s \leftarrow 1$ \\ 59707. else \\ 5971\hspace{3mm}7.1 $r \leftarrow p_0 \mbox{ (mod }8\mbox{)}$ \\ 5972\hspace{3mm}7.2 If $r = 1$ or $r = 7$ then \\ 5973\hspace{6mm}7.2.1 $s \leftarrow 1$ \\ 5974\hspace{3mm}7.3 else \\ 5975\hspace{6mm}7.3.1 $s \leftarrow -1$ \\ 59768. If $p_0 \equiv a'_0 \equiv 3 \mbox{ (mod }4\mbox{)}$ then \\ 5977\hspace{3mm}8.1 $s \leftarrow -s$ \\ 59789. If $a' \ne 1$ then \\ 5979\hspace{3mm}9.1 $p' \leftarrow p \mbox{ (mod }a'\mbox{)}$ \\ 5980\hspace{3mm}9.2 $s \leftarrow s \cdot \mbox{mp\_jacobi}(p', a')$ \\ 598110. $c \leftarrow s$ \\ 598211. Return(\textit{MP\_OKAY}). \\ 5983\hline 5984\end{tabular} 5985\end{center} 5986\end{small} 5987\caption{Algorithm mp\_jacobi} 5988\end{figure} 5989\textbf{Algorithm mp\_jacobi.} 5990This algorithm computes the Jacobi symbol for an arbitrary positive integer $a$ with respect to an odd integer $p$ greater than three. The algorithm 5991is based on algorithm 2.149 of HAC \cite[pp. 73]{HAC}. 5992 5993Step numbers one and two handle the trivial cases of $a = 0$ and $a = 1$ respectively. Step five determines the number of two factors in the 5994input $a$. If $k$ is even than the term $\left ( { 2 \over p } \right )^k$ must always evaluate to one. If $k$ is odd than the term evaluates to one 5995if $p_0$ is congruent to one or seven modulo eight, otherwise it evaluates to $-1$. After the the $\left ( { 2 \over p } \right )^k$ term is handled 5996the $(-1)^{(p-1)(a'-1)/4}$ is computed and multiplied against the current product $s$. The latter term evaluates to one if both $p$ and $a'$ 5997are congruent to one modulo four, otherwise it evaluates to negative one. 5998 5999By step nine if $a'$ does not equal one a recursion is required. Step 9.1 computes $p' \equiv p \mbox{ (mod }a'\mbox{)}$ and will recurse to compute 6000$\left ( {p' \over a'} \right )$ which is multiplied against the current Jacobi product. 6001 6002EXAM,bn_mp_jacobi.c 6003 6004As a matter of practicality the variable $a'$ as per the pseudo-code is reprensented by the variable $a1$ since the $'$ symbol is not valid for a C 6005variable name character. 6006 6007The two simple cases of $a = 0$ and $a = 1$ are handled at the very beginning to simplify the algorithm. If the input is non-trivial the algorithm 6008has to proceed compute the Jacobi. The variable $s$ is used to hold the current Jacobi product. Note that $s$ is merely a C ``int'' data type since 6009the values it may obtain are merely $-1$, $0$ and $1$. 6010 6011After a local copy of $a$ is made all of the factors of two are divided out and the total stored in $k$. Technically only the least significant 6012bit of $k$ is required, however, it makes the algorithm simpler to follow to perform an addition. In practice an exclusive-or and addition have the same 6013processor requirements and neither is faster than the other. 6014 6015Line @59, if@ through @70, }@ determines the value of $\left ( { 2 \over p } \right )^k$. If the least significant bit of $k$ is zero than 6016$k$ is even and the value is one. Otherwise, the value of $s$ depends on which residue class $p$ belongs to modulo eight. The value of 6017$(-1)^{(p-1)(a'-1)/4}$ is compute and multiplied against $s$ on lines @73, if@ through @75, }@. 6018 6019Finally, if $a1$ does not equal one the algorithm must recurse and compute $\left ( {p' \over a'} \right )$. 6020 6021\textit{-- Comment about default $s$ and such...} 6022 6023\section{Modular Inverse} 6024\label{sec:modinv} 6025The modular inverse of a number actually refers to the modular multiplicative inverse. Essentially for any integer $a$ such that $(a, p) = 1$ there 6026exist another integer $b$ such that $ab \equiv 1 \mbox{ (mod }p\mbox{)}$. The integer $b$ is called the multiplicative inverse of $a$ which is 6027denoted as $b = a^{-1}$. Technically speaking modular inversion is a well defined operation for any finite ring or field not just for rings and 6028fields of integers. However, the former will be the matter of discussion. 6029 6030The simplest approach is to compute the algebraic inverse of the input. That is to compute $b \equiv a^{\Phi(p) - 1}$. If $\Phi(p)$ is the 6031order of the multiplicative subgroup modulo $p$ then $b$ must be the multiplicative inverse of $a$. The proof of which is trivial. 6032 6033\begin{equation} 6034ab \equiv a \left (a^{\Phi(p) - 1} \right ) \equiv a^{\Phi(p)} \equiv a^0 \equiv 1 \mbox{ (mod }p\mbox{)} 6035\end{equation} 6036 6037However, as simple as this approach may be it has two serious flaws. It requires that the value of $\Phi(p)$ be known which if $p$ is composite 6038requires all of the prime factors. This approach also is very slow as the size of $p$ grows. 6039 6040A simpler approach is based on the observation that solving for the multiplicative inverse is equivalent to solving the linear 6041Diophantine\footnote{See LeVeque \cite[pp. 40-43]{LeVeque} for more information.} equation. 6042 6043\begin{equation} 6044ab + pq = 1 6045\end{equation} 6046 6047Where $a$, $b$, $p$ and $q$ are all integers. If such a pair of integers $ \left < b, q \right >$ exist than $b$ is the multiplicative inverse of 6048$a$ modulo $p$. The extended Euclidean algorithm (Knuth \cite[pp. 342]{TAOCPV2}) can be used to solve such equations provided $(a, p) = 1$. 6049However, instead of using that algorithm directly a variant known as the binary Extended Euclidean algorithm will be used in its place. The 6050binary approach is very similar to the binary greatest common divisor algorithm except it will produce a full solution to the Diophantine 6051equation. 6052 6053\subsection{General Case} 6054\newpage\begin{figure}[!here] 6055\begin{small} 6056\begin{center} 6057\begin{tabular}{l} 6058\hline Algorithm \textbf{mp\_invmod}. \\ 6059\textbf{Input}. mp\_int $a$ and $b$, $(a, b) = 1$, $p \ge 2$, $0 < a < p$. \\ 6060\textbf{Output}. The modular inverse $c \equiv a^{-1} \mbox{ (mod }b\mbox{)}$. \\ 6061\hline \\ 60621. If $b \le 0$ then return(\textit{MP\_VAL}). \\ 60632. If $b_0 \equiv 1 \mbox{ (mod }2\mbox{)}$ then use algorithm fast\_mp\_invmod. \\ 60643. $x \leftarrow \vert a \vert, y \leftarrow b$ \\ 60654. If $x_0 \equiv y_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ then return(\textit{MP\_VAL}). \\ 60665. $B \leftarrow 0, C \leftarrow 0, A \leftarrow 1, D \leftarrow 1$ \\ 60676. While $u.used > 0$ and $u_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\ 6068\hspace{3mm}6.1 $u \leftarrow \lfloor u / 2 \rfloor$ \\ 6069\hspace{3mm}6.2 If ($A.used > 0$ and $A_0 \equiv 1 \mbox{ (mod }2\mbox{)}$) or ($B.used > 0$ and $B_0 \equiv 1 \mbox{ (mod }2\mbox{)}$) then \\ 6070\hspace{6mm}6.2.1 $A \leftarrow A + y$ \\ 6071\hspace{6mm}6.2.2 $B \leftarrow B - x$ \\ 6072\hspace{3mm}6.3 $A \leftarrow \lfloor A / 2 \rfloor$ \\ 6073\hspace{3mm}6.4 $B \leftarrow \lfloor B / 2 \rfloor$ \\ 60747. While $v.used > 0$ and $v_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\ 6075\hspace{3mm}7.1 $v \leftarrow \lfloor v / 2 \rfloor$ \\ 6076\hspace{3mm}7.2 If ($C.used > 0$ and $C_0 \equiv 1 \mbox{ (mod }2\mbox{)}$) or ($D.used > 0$ and $D_0 \equiv 1 \mbox{ (mod }2\mbox{)}$) then \\ 6077\hspace{6mm}7.2.1 $C \leftarrow C + y$ \\ 6078\hspace{6mm}7.2.2 $D \leftarrow D - x$ \\ 6079\hspace{3mm}7.3 $C \leftarrow \lfloor C / 2 \rfloor$ \\ 6080\hspace{3mm}7.4 $D \leftarrow \lfloor D / 2 \rfloor$ \\ 60818. If $u \ge v$ then \\ 6082\hspace{3mm}8.1 $u \leftarrow u - v$ \\ 6083\hspace{3mm}8.2 $A \leftarrow A - C$ \\ 6084\hspace{3mm}8.3 $B \leftarrow B - D$ \\ 60859. else \\ 6086\hspace{3mm}9.1 $v \leftarrow v - u$ \\ 6087\hspace{3mm}9.2 $C \leftarrow C - A$ \\ 6088\hspace{3mm}9.3 $D \leftarrow D - B$ \\ 608910. If $u \ne 0$ goto step 6. \\ 609011. If $v \ne 1$ return(\textit{MP\_VAL}). \\ 609112. While $C \le 0$ do \\ 6092\hspace{3mm}12.1 $C \leftarrow C + b$ \\ 609313. While $C \ge b$ do \\ 6094\hspace{3mm}13.1 $C \leftarrow C - b$ \\ 609514. $c \leftarrow C$ \\ 609615. Return(\textit{MP\_OKAY}). \\ 6097\hline 6098\end{tabular} 6099\end{center} 6100\end{small} 6101\end{figure} 6102\textbf{Algorithm mp\_invmod.} 6103This algorithm computes the modular multiplicative inverse of an integer $a$ modulo an integer $b$. This algorithm is a variation of the 6104extended binary Euclidean algorithm from HAC \cite[pp. 608]{HAC}. It has been modified to only compute the modular inverse and not a complete 6105Diophantine solution. 6106 6107If $b \le 0$ than the modulus is invalid and MP\_VAL is returned. Similarly if both $a$ and $b$ are even then there cannot be a multiplicative 6108inverse for $a$ and the error is reported. 6109 6110The astute reader will observe that steps seven through nine are very similar to the binary greatest common divisor algorithm mp\_gcd. In this case 6111the other variables to the Diophantine equation are solved. The algorithm terminates when $u = 0$ in which case the solution is 6112 6113\begin{equation} 6114Ca + Db = v 6115\end{equation} 6116 6117If $v$, the greatest common divisor of $a$ and $b$ is not equal to one then the algorithm will report an error as no inverse exists. Otherwise, $C$ 6118is the modular inverse of $a$. The actual value of $C$ is congruent to, but not necessarily equal to, the ideal modular inverse which should lie 6119within $1 \le a^{-1} < b$. Step numbers twelve and thirteen adjust the inverse until it is in range. If the original input $a$ is within $0 < a < p$ 6120then only a couple of additions or subtractions will be required to adjust the inverse. 6121 6122EXAM,bn_mp_invmod.c 6123 6124\subsubsection{Odd Moduli} 6125 6126When the modulus $b$ is odd the variables $A$ and $C$ are fixed and are not required to compute the inverse. In particular by attempting to solve 6127the Diophantine $Cb + Da = 1$ only $B$ and $D$ are required to find the inverse of $a$. 6128 6129The algorithm fast\_mp\_invmod is a direct adaptation of algorithm mp\_invmod with all all steps involving either $A$ or $C$ removed. This 6130optimization will halve the time required to compute the modular inverse. 6131 6132\section{Primality Tests} 6133 6134A non-zero integer $a$ is said to be prime if it is not divisible by any other integer excluding one and itself. For example, $a = 7$ is prime 6135since the integers $2 \ldots 6$ do not evenly divide $a$. By contrast, $a = 6$ is not prime since $a = 6 = 2 \cdot 3$. 6136 6137Prime numbers arise in cryptography considerably as they allow finite fields to be formed. The ability to determine whether an integer is prime or 6138not quickly has been a viable subject in cryptography and number theory for considerable time. The algorithms that will be presented are all 6139probablistic algorithms in that when they report an integer is composite it must be composite. However, when the algorithms report an integer is 6140prime the algorithm may be incorrect. 6141 6142As will be discussed it is possible to limit the probability of error so well that for practical purposes the probablity of error might as 6143well be zero. For the purposes of these discussions let $n$ represent the candidate integer of which the primality is in question. 6144 6145\subsection{Trial Division} 6146 6147Trial division means to attempt to evenly divide a candidate integer by small prime integers. If the candidate can be evenly divided it obviously 6148cannot be prime. By dividing by all primes $1 < p \le \sqrt{n}$ this test can actually prove whether an integer is prime. However, such a test 6149would require a prohibitive amount of time as $n$ grows. 6150 6151Instead of dividing by every prime, a smaller, more mangeable set of primes may be used instead. By performing trial division with only a subset 6152of the primes less than $\sqrt{n} + 1$ the algorithm cannot prove if a candidate is prime. However, often it can prove a candidate is not prime. 6153 6154The benefit of this test is that trial division by small values is fairly efficient. Specially compared to the other algorithms that will be 6155discussed shortly. The probability that this approach correctly identifies a composite candidate when tested with all primes upto $q$ is given by 6156$1 - {1.12 \over ln(q)}$. The graph (\ref{pic:primality}, will be added later) demonstrates the probability of success for the range 6157$3 \le q \le 100$. 6158 6159At approximately $q = 30$ the gain of performing further tests diminishes fairly quickly. At $q = 90$ further testing is generally not going to 6160be of any practical use. In the case of LibTomMath the default limit $q = 256$ was chosen since it is not too high and will eliminate 6161approximately $80\%$ of all candidate integers. The constant \textbf{PRIME\_SIZE} is equal to the number of primes in the test base. The 6162array \_\_prime\_tab is an array of the first \textbf{PRIME\_SIZE} prime numbers. 6163 6164\begin{figure}[!here] 6165\begin{small} 6166\begin{center} 6167\begin{tabular}{l} 6168\hline Algorithm \textbf{mp\_prime\_is\_divisible}. \\ 6169\textbf{Input}. mp\_int $a$ \\ 6170\textbf{Output}. $c = 1$ if $n$ is divisible by a small prime, otherwise $c = 0$. \\ 6171\hline \\ 61721. for $ix$ from $0$ to $PRIME\_SIZE$ do \\ 6173\hspace{3mm}1.1 $d \leftarrow n \mbox{ (mod }\_\_prime\_tab_{ix}\mbox{)}$ \\ 6174\hspace{3mm}1.2 If $d = 0$ then \\ 6175\hspace{6mm}1.2.1 $c \leftarrow 1$ \\ 6176\hspace{6mm}1.2.2 Return(\textit{MP\_OKAY}). \\ 61772. $c \leftarrow 0$ \\ 61783. Return(\textit{MP\_OKAY}). \\ 6179\hline 6180\end{tabular} 6181\end{center} 6182\end{small} 6183\caption{Algorithm mp\_prime\_is\_divisible} 6184\end{figure} 6185\textbf{Algorithm mp\_prime\_is\_divisible.} 6186This algorithm attempts to determine if a candidate integer $n$ is composite by performing trial divisions. 6187 6188EXAM,bn_mp_prime_is_divisible.c 6189 6190The algorithm defaults to a return of $0$ in case an error occurs. The values in the prime table are all specified to be in the range of a 6191mp\_digit. The table \_\_prime\_tab is defined in the following file. 6192 6193EXAM,bn_prime_tab.c 6194 6195Note that there are two possible tables. When an mp\_digit is 7-bits long only the primes upto $127$ may be included, otherwise the primes 6196upto $1619$ are used. Note that the value of \textbf{PRIME\_SIZE} is a constant dependent on the size of a mp\_digit. 6197 6198\subsection{The Fermat Test} 6199The Fermat test is probably one the oldest tests to have a non-trivial probability of success. It is based on the fact that if $n$ is in 6200fact prime then $a^{n} \equiv a \mbox{ (mod }n\mbox{)}$ for all $0 < a < n$. The reason being that if $n$ is prime than the order of 6201the multiplicative sub group is $n - 1$. Any base $a$ must have an order which divides $n - 1$ and as such $a^n$ is equivalent to 6202$a^1 = a$. 6203 6204If $n$ is composite then any given base $a$ does not have to have a period which divides $n - 1$. In which case 6205it is possible that $a^n \nequiv a \mbox{ (mod }n\mbox{)}$. However, this test is not absolute as it is possible that the order 6206of a base will divide $n - 1$ which would then be reported as prime. Such a base yields what is known as a Fermat pseudo-prime. Several 6207integers known as Carmichael numbers will be a pseudo-prime to all valid bases. Fortunately such numbers are extremely rare as $n$ grows 6208in size. 6209 6210\begin{figure}[!here] 6211\begin{small} 6212\begin{center} 6213\begin{tabular}{l} 6214\hline Algorithm \textbf{mp\_prime\_fermat}. \\ 6215\textbf{Input}. mp\_int $a$ and $b$, $a \ge 2$, $0 < b < a$. \\ 6216\textbf{Output}. $c = 1$ if $b^a \equiv b \mbox{ (mod }a\mbox{)}$, otherwise $c = 0$. \\ 6217\hline \\ 62181. $t \leftarrow b^a \mbox{ (mod }a\mbox{)}$ \\ 62192. If $t = b$ then \\ 6220\hspace{3mm}2.1 $c = 1$ \\ 62213. else \\ 6222\hspace{3mm}3.1 $c = 0$ \\ 62234. Return(\textit{MP\_OKAY}). \\ 6224\hline 6225\end{tabular} 6226\end{center} 6227\end{small} 6228\caption{Algorithm mp\_prime\_fermat} 6229\end{figure} 6230\textbf{Algorithm mp\_prime\_fermat.} 6231This algorithm determines whether an mp\_int $a$ is a Fermat prime to the base $b$ or not. It uses a single modular exponentiation to 6232determine the result. 6233 6234EXAM,bn_mp_prime_fermat.c 6235 6236\subsection{The Miller-Rabin Test} 6237The Miller-Rabin (citation) test is another primality test which has tighter error bounds than the Fermat test specifically with sequentially chosen 6238candidate integers. The algorithm is based on the observation that if $n - 1 = 2^kr$ and if $b^r \nequiv \pm 1$ then after upto $k - 1$ squarings the 6239value must be equal to $-1$. The squarings are stopped as soon as $-1$ is observed. If the value of $1$ is observed first it means that 6240some value not congruent to $\pm 1$ when squared equals one which cannot occur if $n$ is prime. 6241 6242\begin{figure}[!here] 6243\begin{small} 6244\begin{center} 6245\begin{tabular}{l} 6246\hline Algorithm \textbf{mp\_prime\_miller\_rabin}. \\ 6247\textbf{Input}. mp\_int $a$ and $b$, $a \ge 2$, $0 < b < a$. \\ 6248\textbf{Output}. $c = 1$ if $a$ is a Miller-Rabin prime to the base $a$, otherwise $c = 0$. \\ 6249\hline 62501. $a' \leftarrow a - 1$ \\ 62512. $r \leftarrow n1$ \\ 62523. $c \leftarrow 0, s \leftarrow 0$ \\ 62534. While $r.used > 0$ and $r_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\ 6254\hspace{3mm}4.1 $s \leftarrow s + 1$ \\ 6255\hspace{3mm}4.2 $r \leftarrow \lfloor r / 2 \rfloor$ \\ 62565. $y \leftarrow b^r \mbox{ (mod }a\mbox{)}$ \\ 62576. If $y \nequiv \pm 1$ then \\ 6258\hspace{3mm}6.1 $j \leftarrow 1$ \\ 6259\hspace{3mm}6.2 While $j \le (s - 1)$ and $y \nequiv a'$ \\ 6260\hspace{6mm}6.2.1 $y \leftarrow y^2 \mbox{ (mod }a\mbox{)}$ \\ 6261\hspace{6mm}6.2.2 If $y = 1$ then goto step 8. \\ 6262\hspace{6mm}6.2.3 $j \leftarrow j + 1$ \\ 6263\hspace{3mm}6.3 If $y \nequiv a'$ goto step 8. \\ 62647. $c \leftarrow 1$\\ 62658. Return(\textit{MP\_OKAY}). \\ 6266\hline 6267\end{tabular} 6268\end{center} 6269\end{small} 6270\caption{Algorithm mp\_prime\_miller\_rabin} 6271\end{figure} 6272\textbf{Algorithm mp\_prime\_miller\_rabin.} 6273This algorithm performs one trial round of the Miller-Rabin algorithm to the base $b$. It will set $c = 1$ if the algorithm cannot determine 6274if $b$ is composite or $c = 0$ if $b$ is provably composite. The values of $s$ and $r$ are computed such that $a' = a - 1 = 2^sr$. 6275 6276If the value $y \equiv b^r$ is congruent to $\pm 1$ then the algorithm cannot prove if $a$ is composite or not. Otherwise, the algorithm will 6277square $y$ upto $s - 1$ times stopping only when $y \equiv -1$. If $y^2 \equiv 1$ and $y \nequiv \pm 1$ then the algorithm can report that $a$ 6278is provably composite. If the algorithm performs $s - 1$ squarings and $y \nequiv -1$ then $a$ is provably composite. If $a$ is not provably 6279composite then it is \textit{probably} prime. 6280 6281EXAM,bn_mp_prime_miller_rabin.c 6282 6283 6284 6285 6286\backmatter 6287\appendix 6288\begin{thebibliography}{ABCDEF} 6289\bibitem[1]{TAOCPV2} 6290Donald Knuth, \textit{The Art of Computer Programming}, Third Edition, Volume Two, Seminumerical Algorithms, Addison-Wesley, 1998 6291 6292\bibitem[2]{HAC} 6293A. Menezes, P. van Oorschot, S. Vanstone, \textit{Handbook of Applied Cryptography}, CRC Press, 1996 6294 6295\bibitem[3]{ROSE} 6296Michael Rosing, \textit{Implementing Elliptic Curve Cryptography}, Manning Publications, 1999 6297 6298\bibitem[4]{COMBA} 6299Paul G. Comba, \textit{Exponentiation Cryptosystems on the IBM PC}. IBM Systems Journal 29(4): 526-538 (1990) 6300 6301\bibitem[5]{KARA} 6302A. Karatsuba, Doklay Akad. Nauk SSSR 145 (1962), pp.293-294 6303 6304\bibitem[6]{KARAP} 6305Andre Weimerskirch and Christof Paar, \textit{Generalizations of the Karatsuba Algorithm for Polynomial Multiplication}, Submitted to Design, Codes and Cryptography, March 2002 6306 6307\bibitem[7]{BARRETT} 6308Paul Barrett, \textit{Implementing the Rivest Shamir and Adleman Public Key Encryption Algorithm on a Standard Digital Signal Processor}, Advances in Cryptology, Crypto '86, Springer-Verlag. 6309 6310\bibitem[8]{MONT} 6311P.L.Montgomery. \textit{Modular multiplication without trial division}. Mathematics of Computation, 44(170):519-521, April 1985. 6312 6313\bibitem[9]{DRMET} 6314Chae Hoon Lim and Pil Joong Lee, \textit{Generating Efficient Primes for Discrete Log Cryptosystems}, POSTECH Information Research Laboratories 6315 6316\bibitem[10]{MMB} 6317J. Daemen and R. Govaerts and J. Vandewalle, \textit{Block ciphers based on Modular Arithmetic}, State and {P}rogress in the {R}esearch of {C}ryptography, 1993, pp. 80-89 6318 6319\bibitem[11]{RSAREF} 6320R.L. Rivest, A. Shamir, L. Adleman, \textit{A Method for Obtaining Digital Signatures and Public-Key Cryptosystems} 6321 6322\bibitem[12]{DHREF} 6323Whitfield Diffie, Martin E. Hellman, \textit{New Directions in Cryptography}, IEEE Transactions on Information Theory, 1976 6324 6325\bibitem[13]{IEEE} 6326IEEE Standard for Binary Floating-Point Arithmetic (ANSI/IEEE Std 754-1985) 6327 6328\bibitem[14]{GMP} 6329GNU Multiple Precision (GMP), \url{http://www.swox.com/gmp/} 6330 6331\bibitem[15]{MPI} 6332Multiple Precision Integer Library (MPI), Michael Fromberger, \url{http://thayer.dartmouth.edu/~sting/mpi/} 6333 6334\bibitem[16]{OPENSSL} 6335OpenSSL Cryptographic Toolkit, \url{http://openssl.org} 6336 6337\bibitem[17]{LIP} 6338Large Integer Package, \url{http://home.hetnet.nl/~ecstr/LIP.zip} 6339 6340\bibitem[18]{ISOC} 6341JTC1/SC22/WG14, ISO/IEC 9899:1999, ``A draft rationale for the C99 standard.'' 6342 6343\bibitem[19]{JAVA} 6344The Sun Java Website, \url{http://java.sun.com/} 6345 6346\end{thebibliography} 6347 6348\input{tommath.ind} 6349 6350\end{document} 6351